Homework answers IEE1149 Summer 2015 - Social Sciences2. Read \Which Price is Right?" by Charles...

$: Homework answers IEE1149 Summer 2015 - Social Sciences2. Read \Which Price is Right?" by Charles Fishman (available at the \Media Clips" link). The article mentions a behavior which$
Homework answers IEE1149 Summer 2015

1. Read “You Yawn, We All Yawn—And Empathy May Explain Why” by Alison McCook(available at the “Media Clips” link on the course website). Is the explanation here of whya person yawns a rational choice explanation? Why or why not?I would say that this is not a rational choice explanation because it does not assume that aperson makes a choice of whether to yawn or not.

2. Read “Which Price is Right?” by Charles Fishman (available at the “Media Clips”link). The article mentions a behavior which seems to be a violation of our simple model ofindividual choice as we defined it in class. Point out the behavior and explain mathematicallywhy a rational choice model (at least the most obvious one) cannot capture this behavior.What do you think explains the behavior?The article says that if there are two versions of a product, one which sells for $14.95 anda more glitzy one which sells for $18.95, then a person typically chooses the $14.95 one.However, if a $34.95 version is introduced, then a person is more likely to buy the $18.95version.The most obvious rational choice model of this assigns a payoff to each of the three alterna-tives. Since a typical person buys the $14.95 version over the $18.95 version, the payoff frombuying the $14.95 version (i.e. considering the price and how good the product is) must behigher than the payoff from the $18.95 version. But when the $34.95 version is introduced,the typical person buys the $18.95 version, which means that the payoff from the $18.95version must be higher than the payoff from the $14.95 version (and the payoff from the$34.95 version). This is a contradiction. Hence a rational choice model cannot explain thisbehavior.The standard interpretation of this kind of effect is “framing”: when there are only twoversions available, a person thinks she is wasting money buying the one which is moreexpensive but with the same functionality. However, if an even more expensive version isavailable, then a person thinks that he is “reasonable” if he buys the middle-priced product.For example, it seems that many stores (like stereo and video stores) have very expensivestuff on display which almost no one buys, and perhaps the reason is that these items makethe less expensive (but still fairly pricey) stuff seem reasonable.The interpretation of this which is more rational-choice-like in spirit says that there is alwayssome uncertainty about the quality of a product, and that the relative price of a given productand its position in a product line provides some information about it. For example, a personmight think that if a $34.95 version of the product is available, quality of the product mustbe an important issue, and hence one shouldn’t buy the absolute lowest quality product.Another interpretation is that perhaps a person always uses the rule “Buy the middle ofthe line product” not because it always is the best choice, but because it saves time andheadaches, and that this strategy averaged over one’s entire life is the best one.

3. Read “Does Blanket ‘Don’t Go to Graduate School!’ Advice Ignore Race and Reality?”by Tressie McMillan Cottom (available at the “Media Clips” link). Can you make McMillanCottom’s argument using our simple model of individual choice? Or is McMillan Cottommaking a different kind of argument?I think that Cottom’s argument can be understood using our simple model of individualchoice. It’s like the example discussed in class of a person who won’t protest if the costof protesting is too high, but will protest even when costs are high if she just can’t standstaying at home and doing nothing.Say that if you could go to graduate school for free, you would get a utility of 20. Butgraduate school costs 8. If you come from a middle class family and can thus can count ona lifestyle with utility 15 with just a bachelor’s degree, then paying to go to graduate schooldoesn’t make sense, because the net utility of going to graduate school would be 20−8 = 12,which is less than 15. But if you come from a working family and expect a lifestyle ofutility 10 with just a bachelor’s degree, then paying to go to graduate school does makesense, because 12 is greater than 10. In other words, for the person from a working family,graduate school moves you farther (20 − 10 = 10) than for a person from a middle-classfamily (20− 15 = 5).

4. Say that you and a friend are meeting for lunch. Both you and your friend can either belate or on time. If both of you are on time, you each get a utility of 3. If one is on timeand the other is late, the prompt one gets a utility of 1 (since she has to wait around doingnothing) and the tardy one gets a utility of 4 (since she doesn’t have to wait). However, ifboth are late, you don’t find each other and you each get a utility of 0.

a. Model this as a strategic form game.The game looks like this.

Friend is late Friend is on time

You are late 0,0 4,1

You are on time 1,4 3,3

This game is the same as the chicken game discussed in class, where late is “straight” andon time is “swerve.”

5. Say you and a friend each privately choose a whole number between 0 and 5 (that is: 0, 1,2, 3, 4, or 5). If you both choose the same number, I will give you both that number times$100. If your number is exactly one less than your opponent’s, however, you will get youropponent’s number times $100 plus a bonus $100 and your opponent will get nothing. Inany other case, both of you get nothing. So for example, if you both choose the number 5, Iwill give you both $500. If you choose 4 and your opponent chooses 5, you will get $600 andyour opponent nothing. If you choose 3 and your opponent chooses 5, you both get nothing.

a. Model this as a strategic form game.This game looks like this:

2 chooses 0 2 chooses 1 2 chooses 2 2 chooses 3 2 chooses 4 2 chooses 5

1 chooses 0 0, 0 200, 0 0, 0 0, 0 0, 0 0, 0

1 chooses 1 0, 200 100, 100 300, 0 0, 0 0, 0 0, 0

1 chooses 2 0, 0 0, 300 200, 200 400, 0 0, 0 0, 0

1 chooses 3 0, 0 0, 0 0, 400 300, 300 500, 0 0, 0

1 chooses 4 0, 0 0, 0 0, 0 0, 500 400, 400 600, 0

1 chooses 5 0, 0 0, 0 0, 0 0, 0 0, 600 500, 500

Undercut your opponent

b. Read the article “Hollywood’s Death Spiral” by Edward Jay Epstein on the web site. Canyou use this game to think about the situation described in the article?The game corresponds roughly to the situation described in the article in which each studiowants to release its DVDs slightly sooner than others. Getting your DVD on store shelvessooner is a competitive advantage. However, the sooner your DVD comes out, the less likelypeople will come to the theatrical release, since they know that the DVD will be out soon.As studios release their DVDs earlier and earlier, the industry as a whole loses profits.

6. Ann and Bob are each trying to win a prize in a school raffle (lottery). Each can buyeither 0, 1, 2, or 3 raffle tickets. Ann and Bob are the only two people in the raffle, and eachticket has an equal chance of winning. So for example, if Ann buys 2 tickets and Bob buys3 tickets, then Ann has a 2/5 chance of winning and Bob has a 3/5 chance of winning (if noone buys any tickets, the raffle is cancelled). The prize is worth $60, and both Ann and Bobcare about their “expected payoffs”: for example, if Ann has a 2/5 chance of winning, herexpected payoff is $24. Model the following situations with strategic form games.

a. Say that raffle tickets are free. What does the game look like?We simply do some expected value calculations to derive the following:

2 buys 0 tickets 2 buys 1 ticket 2 buys 2 tickets 2 buys 3 tickets

1 buys 0 tickets 0, 0 0, 60 0, 60 0, 60

1 buys 1 ticket 60, 0 30, 30 20, 40 15, 45

1 buys 2 tickets 60, 0 40, 20 30, 30 24, 36

1 buys 3 tickets 60, 0 45, 15 36, 24 30, 30

Tickets are free

b. Now say that raffle tickets cost $6 each. What does the game look like?2 buys 0 tickets 2 buys 1 ticket 2 buys 2 tickets 2 buys 3 tickets

1 buys 0 tickets 0, 0 0, 54 0, 48 0, 42

1 buys 1 ticket 54, 0 24, 24 14, 28 9, 27

1 buys 2 tickets 48, 0 28, 14 18, 18 12, 18

1 buys 3 tickets 42, 0 27, 9 18, 12 12, 12

Tickets cost $6 each

c. Now say that raffle tickets cost $10 each. What does the game look like?

Now the game looks like this:2 buys 0 tickets 2 buys 1 ticket 2 buys 2 tickets 2 buys 3 tickets

1 buys 0 tickets 0, 0 0, 50 0, 40 0, 30

1 buys 1 ticket 50, 0 20, 20 10, 20 5, 15

1 buys 2 tickets 40, 0 20, 10 10, 10 4, 6

1 buys 3 tickets 30, 0 15, 5 6, 4 0, 0

Tickets cost $10 each

7. Say that Spy 1 is trying to listen in on Spy 2. There are three rooms, A, B, and C,arranged in a line like this: A—B—C. In other words, A is on the left, B is in the middle,and C is on the right. Each spy must decide independently and simultaneously which roomto enter. Their payoffs are determined as follows. If they both choose the same room, thenthey will see each other, a bloody gun battle will ensue, and both get payoff −10. If theyare in adjacent rooms (for example, if Spy 1 is in room A and Spy 2 is in room B) then Spy1 can set up her eavesdropping equipment and can intercept Spy 2’s communications; henceSpy 1 gets a payoff of 5 and Spy 2 gets a payoff of −5. If they are not in adjacent rooms andthey are not in the same room (for example, if Spy 1 is in room A and Spy 2 is in room C)then the distance between them is too great for the eavesdropping equipment to work; Spy1 gets no secrets and Spy 2 gets to keep hers, and so both get a payoff of 0.

a. Model this as a strategic form game.The game looks like this:

2A 2B 2C

1A −10,−10 5,−5 0, 0

1B 5,−5 −10,−10 5,−5

1C 0, 0 5,−5 −10,−10

8. Say that persons 1, 2, and 3 each decide whether to go to restaurant A or restaurant B.Person 1 wants the dinner group to be as large as possible. For person 1, the worst thingis if she goes to a restaurant alone, the best thing is if all three go to the same place, andgoing with one person (it doesn’t matter which) is OK, neither best or worst. Person 2 isthe exact opposite; she wants the dinner group to be as small as possible. All person 3 caresabout is going to the same place as person 1, since he likes person 1.

a. Model this as a strategic form game.Each person can go to A or go to B. The game looks like this.

2 goes to A 2 goes to B

1 goes to A 10, 0, 10 5, 10, 10

1 goes to B 0, 5, 0 5, 5, 0

3 goes to A


1 goes to A 5, 5, 0 0, 5, 0

1 goes to B 5, 10, 10 10, 0, 10

3 goes to B

b. Now say that person 3 loses interest in person 1 and becomes grouchy like person 2. Modelthis as a strategic form game.Now the game looks like this.


1 goes to A 10, 0, 0 5, 10, 5

1 goes to B 0, 5, 5 5, 5, 10

3 goes to A


1 goes to A 5, 5, 10 0, 5, 5

1 goes to B 5, 10, 5 10, 0, 0

3 goes to B

9. Say that there are two people, a security guard and a thief. The security guard can eitherbe vigilant or relax. The thief can either steal or do nothing. If the guard is vigilant, thenthe thief would rather do nothing than steal. If the guard is relaxed, however, the thiefwould rather steal than do nothing. If the thief steals, the guard would rather be vigilantthan be relaxed. If the thief does nothing, however, the guard would rather be relaxed thanvigilant.

a. Model this as a strategic form game. Feel free to choose payoffs which make sense to you.My version of this game looks like this:

thief steals thief does nothing

guard is vigilant 5,-10 -2,0

guard is relaxed -10,10 0,0

10. Read “Running Mates: The Clark-Lieberman Iowa Jailbreak” by William Saletan (atthe “Media Clips” link). Model the situation as a strategic form game.Wesley Clark and Joe Lieberman can each decide whether to stay in the Iowa caucusesor quit. The article argues that the worst thing for either candidate is to be the only onequitting Iowa, because it makes that candidate look like a loser. If Clark quits and Liebermanstays in, Lieberman gets a slightly higher payoff than if both stayed in (since Lieberman getssome of Clark’s voters), and similarly, Clark gets a slightly higher payoff if only Liebermandrops out. The best thing for both candidates, however, is for both to quit, so they canconcentrate on the upcoming New Hampshire primary. So this is the game I come up with:

Lieberman stays Lieberman quits

Clark stays 0,0 1,-10

Clark quits -10,1 5,5

11. In the movie Return to Paradise (see the “Media Clips” link), Sheriff, Tony, and Lewiswent to Malaysia and did various illegal things. After Sheriff and Tony left, Lewis wascharged and scheduled to be executed. If either Sheriff or Tony returns to Malaysia andadmits shared responsibility, Lewis’s sentence will be reduced and he will live. If bothSheriff and Tony return, then each will have to serve three years in prison in Malaysia. Ifonly one returns, then that person will have to serve six years in prison. The two players,Sheriff and Tony, can each either decide to go back to Malaysia or stay in New York. Modelthe situation as a game with two players (Sheriff and Tony). Feel free to choose payoffswhich make sense to you (that’s what makes the problem kind of interesting).The two players, Sheriff and Tony, can each either decide to go back to Malaysia or stay inNew York. What you think the game is depends on your assumptions about how much guiltSheriff and Tony will feel.For example, say that Sheriff and Tony care only about avoiding jail themselves, and allother things equal would be happy to have Lewis alive. Then the game would look like this:

Tony stays in NYC Tony goes back to Malaysia

Sheriff stays in NYC -1,-1 -0,-6

Sheriff goes back to Malaysia -6,0 -3,-3

In other words, the best thing would be to have the other guy go back and you stay at home.Here you can think of Lewis’s life as worth 1 year of jail time to Sheriff and Tony.Another possibility is that Lewis’s life is worth 4 years of jail time—in other words, neitherwill go back if he has go back alone. Then the game would look like this:




Note that this game is a Prisoners’ Dilemma: if you know the other person is going back,you have the incentive to not get on the plane and stay in NYC; however, if you both stayin NYC, Lewis dies and youre worse off than if you both returned to Malaysia.Another possibility is that Lewis’s life is worth 10 years of jail time—having Lewis die is theabsolute worst thing. Then the game would look like this:




Note that this game is a game of chicken, where Lewis dying is like the two cars crashing intoeach other. Still the best thing for each person is to stay home and have the other personto return to Malaysia.Finally, say that people feel really guilty and basically internalize the others’ pain. So if youTony goes back to Malaysia and Sheriff doesn’t, Sheriff feels just as bad as Tony.



Sheriff goes back to Malaysia -6,-6 -3,-3

12. Say that you and a friend are meeting for lunch. Both you and your friend can eitherbe late or on time. If both of you are on time, you each get a utility of 3. If one is on timeand the other is late, the prompt one gets a utility of 1 (since she has to wait around doingnothing) and the tardy one gets a utility of 4 (since she doesn’t have to wait). However, ifboth are late, you don’t find each other and you each get a utility of 0.

a. Model this as a strategic form game and find all (pure strategy and mixed strategy) Nashequilibria.The game looks like

On time Late

On time 3,3 1,4

Late 4,1 0,0

b. Are there strongly or weakly dominated strategies in this game?There are no strongly or weakly dominated strategies in this game.

c. Find the (pure strategy) Nash equilibria of this game.The pure strategy Nash equilibria are (Late, On time) and (On time, late).

13. Say you and a friend each privately choose a whole number between 0 and 5 (that is: 0,1, 2, 3, 4, or 5). If you both choose the same number, I will give you both that number times$100. If your number is exactly one less than your opponent’s, however, you will get youropponent’s number times $100 plus a bonus $100 and your opponent will get nothing. Inany other case, both of you get nothing. So for example, if you both choose the number 5, Iwill give you both $500. If you choose 4 and your opponent chooses 5, you will get $600 andyour opponent nothing. If you choose 3 and your opponent chooses 5, you both get nothing.Model this as a strategic form game.

a. “Solve” this game by iteratively eliminating weakly dominated strategies.This game looks like this:

2 chooses 0 2 chooses 1 2 chooses 2 2 chooses 3 2 chooses 4 2 chooses 5

1 chooses 0 0, 0 200, 0 0, 0 0, 0 0, 0 0, 0

1 chooses 1 0, 200 100, 100 300, 0 0, 0 0, 0 0, 0

1 chooses 2 0, 0 0, 300 200, 200 400, 0 0, 0 0, 0

1 chooses 3 0, 0 0, 0 0, 400 300, 300 500, 0 0, 0

1 chooses 4 0, 0 0, 0 0, 0 0, 500 400, 400 600, 0

1 chooses 5 0, 0 0, 0 0, 0 0, 0 0, 600 500, 500

Undercut your opponentNote that for player 1, choosing 4 weakly dominates choosing 5. The only way player 1makes money by choosing 5 is if player 2 also chooses 5. But then player 1 is better offchoosing 4. Choosing 4 is also better if player 2 chooses 4 also. So choosing 4 is always atleast as good as choosing 4. Once we eliminate this strategy, we can use the same reasoningto conclude that player 2 should not choose 5 as well. Then we get the following game:

2 chooses 0 2 chooses 1 2 chooses 2 2 chooses 3 2 chooses 4

1 chooses 0 0, 0 200, 0 0, 0 0, 0 0, 0

1 chooses 1 0, 200 100, 100 300, 0 0, 0 0, 0

1 chooses 2 0, 0 0, 300 200, 200 400, 0 0, 0

1 chooses 3 0, 0 0, 0 0, 400 300, 300 500, 0

1 chooses 4 0, 0 0, 0 0, 0 0, 500 400, 400

After strategies have been eliminatedThen we can use similar reasoning to conclude that for both players, choosing 3 weaklydominates choosing 4. We can continue to eliminate strategies successively until we get

2 chooses 0 2 chooses 1

1 chooses 0 0, 0 200, 0

1 chooses 1 0, 200 100, 100

What’s leftNow we can say that choosing 0 weakly dominates choosing 1 for both players, so all we haveleft is that both players will choose 0. Or, if we eliminate player 1’s strategy of choosing 1first, then we get:

2 chooses 0 2 chooses 1

1 chooses 0 0, 0 200, 0

Now we cannot eliminate player 2’s choosing 2. So we have strategy profiles (0, 0) and (0,1) as what is left. Similarly, if we eliminated player 2’s strategy of choosing 2 first, we wouldget (0, 0) and (1, 0) as the strategy profiles left. This example shows how our result dependson the order in which we eliminated the weakly dominated strategies.

b. Find the (pure strategy) Nash equilibria of this game.If we find the Nash equilibria of the original game, we find that (0, 0), (1, 0) and (0, 1) arethe Nash equilibria. So we can say that the most that I will have to shell out is $100, whichis surprising since the two players could possibly make much more. Since both players havethe incentive to undercut each other, they undercut each other (almost) all the way down.

14. Ann and Bob are each trying to win a prize in a school raffle (lottery). Each can buyeither 0, 1, 2, or 3 raffle tickets. Ann and Bob are the only two people in the raffle, and eachticket has an equal chance of winning. So for example, if Ann buys 2 tickets and Bob buys3 tickets, then Ann has a 2/5 chance of winning and Bob has a 3/5 chance of winning (ifno one buys any tickets, the raffle is cancelled). The prize is worth $60, and both Ann andBob are risk-neutral: for example, if Ann has a 2/5 chance of winning, her expected utilityis $24. Model the following situations with strategic form games.

a. Say that raffle tickets are free. What does the game look like? Are any strategies inthis game strongly or weakly dominated? Can you iteratively eliminate strongly or weaklydominated strategies? What are the (pure strategy) Nash equilibria of this game?We simply do some expected value calculations to derive the following:


1 buys 0 tickets 0, 0 0, 60 0, 60 0, 60

1 buys 1 ticket 60, 0 30, 30 20, 40 15, 45

1 buys 2 tickets 60, 0 40, 20 30, 30 24, 36

1 buys 3 tickets 60, 0 45, 15 36, 24 30, 30

Tickets are freeHere it is easy to see that for both players, the strategy of buying 0 tickets is stronglydominated by any other strategy. Also, for both players, the strategy of buying 1 ticketand the strategy of buying 2 tickets are both weakly dominated by the strategy of buying 3tickets—this makes sense, since if tickets don’t cost anything, there is no reason not to buya lot. It is easy to see that the only pure strategy Nash equilibrium is where player 1 buys3 tickets and player 2 buys 3 tickets.

b. Now say that raffle tickets cost $6 each. What does the game look like? Are any strategiesin this game strongly or weakly dominated? Can you iteratively eliminate strongly or weaklydominated strategies? What are the (pure strategy) Nash equilibria of this game?Now the game looks like this:


1 buys 0 tickets 0, 0 0, 54 0, 48 0, 42

1 buys 1 ticket 54, 0 24, 24 14, 28 9, 27

1 buys 2 tickets 48, 0 28, 14 18, 18 12, 18

1 buys 3 tickets 42, 0 27, 9 18, 12 12, 12

Tickets cost $6 eachNow, again for both players, the strategy of buying 0 tickets is strongly dominated by anyother strategy. But now for both players, the strategy of buying 3 tickets is weakly dominatedby the strategy of buying 2 tickets. There are four pure strategy Nash equilibria: (buy 2tickets, buy 2 tickets), (buy 3 tickets, buy 2 tickets), (buy 2 tickets, buy 3 tickets), and (buy3 tickets, buy 3 tickets).

c. Now say that raffle tickets cost $10 each. What does the game look like? Are any strategiesin this game strongly or weakly dominated? Can you iteratively eliminate strongly or weaklydominated strategies? What are the (pure strategy) Nash equilibria of this game?Now the game looks like this:


1 buys 0 tickets 0, 0 0, 50 0, 40 0, 30

1 buys 1 ticket 50, 0 20, 20 10, 20 5, 15

1 buys 2 tickets 40, 0 20, 10 10, 10 4, 6

1 buys 3 tickets 30, 0 15, 5 6, 4 0, 0

Tickets cost $10 eachNow for both players, the strategy of buying 0 tickets is strongly dominated by the strategyof buying 1 ticket or the strategy of buying 2 tickets. Also, for both players the strategy ofbuying 3 tickets is strongly dominated by the strategy of buying 1 ticket. For both players,the strategy of buying 2 tickets is weakly dominated by the strategy of buying 1 ticket.There are four pure strategy Nash equilibria: (buy 1 ticket, buy 1 ticket), (buy 1 tickets,buy 2 tickets), (buy 2 tickets, buy 1 tickets), and (buy 2 tickets, buy 2 tickets).

15. Say that Spy 1 is trying to listen in on Spy 2. There are three rooms, A, B, and C,arranged in a line like this: A—B—C. In other words, A is on the left, B is in the middle,and C is on the right. Each spy must decide independently and simultaneously which roomto enter. Their payoffs are determined as follows. If they both choose the same room, thenthey will see each other, a bloody gun battle will ensue, and both get payoff −10. If theyare in adjacent rooms (for example, if Spy 1 is in room A and Spy 2 is in room B) then Spy1 can set up her eavesdropping equipment and can intercept Spy 2’s communications; henceSpy 1 gets a payoff of 5 and Spy 2 gets a payoff of −5. If they are not in adjacent rooms andthey are not in the same room (for example, if Spy 1 is in room A and Spy 2 is in room C)then the distance between them is too great for the eavesdropping equipment to work; Spy1 gets no secrets and Spy 2 gets to keep hers, and so both get a payoff of 0.

a. Model this as a strategic form game.The game looks like this:

2A 2B 2C

1A −10,−10 5,−5 0, 0

1B 5,−5 −10,−10 5,−5

1C 0, 0 5,−5 −10,−10

b. Are any strategies in this game strongly or weakly dominated? Can you iterativelyeliminate strongly or weakly dominated strategies? What are the (pure strategy) Nashequilibria of this game?No strategies are strongly or weakly dominated. It is not hard to see that (1B, 2A) and(1B, 2C) are the Nash equilibria of this game. This makes sense: if Spy 1 chooses roomB, Spy 2 has “no choice” but to choose either room A or room C, thus guaranteeing Spy 1eavesdropping access.

16. Say that persons 1, 2, and 3 each decide whether to go to restaurant A or restaurant B.Person 1 wants the dinner group to be as large as possible. For person 1, the worst thingis if she goes to a restaurant alone, the best thing is if all three go to the same place, andgoing with one person (it doesn’t matter which) is OK, neither best or worst. Person 2 isthe exact opposite; she wants the dinner group to be as small as possible. All person 3 caresabout is going to the same place as person 1, since he likes person 1.

a. Model this as a strategic form game.Each person can go to A or go to B. The game looks like this.


1 goes to A 10, 0, 10 5, 10, 10

1 goes to B 0, 5, 0 5, 5, 0

3 goes to A


1 goes to A 5, 5, 0 0, 5, 0

1 goes to B 5, 10, 10 10, 0, 10

3 goes to B

b. Are any strategies in this game strongly or weakly dominated? Can youiteratively eliminate strongly or weakly dominated strategies? What are the(pure strategy) Nash equilibria of this game?In this game, no strategy is weakly or strongly dominated. The two pure strategy Nashequilibria are (1A, 2B, 3A) and (1B, 2A, 3B). Since person 3 likes person 1, he will alwaysgo to where person 1 goes, and person 2 will avoid them both.

c. Now say that person 3 loses interest in person 1 and becomes grouchy like person 2. Modelthis as a strategic form game.Now the game looks like this.


1 goes to A 10, 0, 0 5, 10, 5

1 goes to B 0, 5, 5 5, 5, 10

3 goes to A


1 goes to A 5, 5, 10 0, 5, 5

1 goes to B 5, 10, 5 10, 0, 0

3 goes to B

d. Are any strategies in this game strongly or weakly dominated? Can youiteratively eliminate strongly or weakly dominated strategies? What are the(pure strategy) Nash equilibria of this game?

Again, no strategy is weakly or strongly dominated. There are four pure strategy Nashequilibria: (1A, 2B, 3A), (1B, 2B, 3A), (1A, 2A, 3B), (1B, 2A, 3B). Since it is not possiblefor both grouchy people to be alone, one of them has to eat with someone else. Note thatthe grouchy people will never eat together with person 1 at the other restaurant, becausethen person 1, who likes a big crowd, would prefer to join them. Of course, everyone eatingtogether is not an equilibrium because one of the grouchy people would like to deviate.

17. Say that there are two people, a security guard and a thief. The security guard caneither be vigilant or relax. The thief can either steal or do nothing. If the guard is vigilant,then the thief would rather do nothing than steal. If the guard is relaxed, however, the thiefwould rather steal than do nothing. If the thief steals, the guard would rather be vigilantthan be relaxed. If the thief does nothing, however, the guard would rather be relaxed thanvigilant.

a. Model this as a strategic form game. Feel free to choose payoffs which make sense to you.My version of this game looks like this:

thief steals thief does nothing

guard is vigilant 5,-10 -2,0

guard is relaxed -10,10 0,0

b. What are the (pure strategy) Nash equilibria of this game?In this game, there are no pure strategy Nash equilibria: from every outcome, someone wouldlike to deviate.

18. Read “Running Mates: The Clark-Lieberman Iowa Jailbreak” by William Saletan (atthe “Media Clips” link). Model the situation as a strategic form game.

a. Are any strategies in this game strongly or weakly dominated? Can you iteratively elimi-nate strongly or weakly dominated strategies? What are the (pure strategy) Nash equilibriaof this game?

Lieberman stays Lieberman quits

Clark stays 0,0 1,-10

Clark quits -10,1 5,5

In this game, no strategy is strongly or weakly dominated. There are two Nash equilibria:(Clark stays, Lieberman stays), and (Clark quits, Lieberman quits).

19. Look at Clip 1 of Return to Paradise on the web site (look under the “Media Clips”link). Model the situation as a game with two players (Sheriff and Tony). Feel free to choosepayoffs which make sense to you (that’s what makes the problem kind of interesting).

a. In your version of the game, are there strongly or weakly dominated strategies? What arethe (pure strategy) Nash equilibria of this game?

For example, say that Sheriff and Tony care only about avoiding jail themselves, and allother things equal would be happy to have Lewis alive. Then the game would look like this:


Sheriff stays in NYC -1,-1 0,-6


In other words, the best thing would be to have the other guy go back and you stay at home.Here you can think of Lewis’s life as worth 1 year of jail time to Sheriff and Tony.In this game, staying in NYC strongly dominates going back to Malaysia for both people.The only Nash equilibrium is (NYC, NYC).

Another possibility is that Lewis’s life is worth 4 years of jail time—in other words, neitherwill go back if he has go go back alone. Then the game would look like this:




Note that this game is a Prisoners’ Dilemma: if you know the other person is going back,you have the incentive to not get on the plane and stay in NYC; however, if you both stayin NYC, Lewis dies and you’re worse off than if you both returned to Malaysia. Here, againNYC strongly dominates Malaysia and the only Nash equilibrium is (NYC, NYC).

Another possibility is that Lewis’s life is worth 10 years of jail time—having Lewis die is theabsolute worst thing. Then the game would look like this:




Note that this game is a game of chicken, where Lewis dying is like the two cars crashing intoeach other. Still the best thing for each person is to stay home and have the other personto return to Malaysia. Here, no strategy is weakly or strongly dominated and there are twoNash equilibria: (NYC, Malaysia) and (Malaysia, NYC).

Finally, say that people feel really guilty and basically internalize the others’ pain. So if youTony goes back to Malaysia and Sheriff doesn’t, Sheriff feels just as bad as Tony.



Sheriff goes back to Malaysia -6,-6 -3,-3

Now Malaysia strongly dominates NYC for both players, and the only Nash equilibrium is(Malaysia, Malaysia).

20. Read “Report Calls Recycling Costlier Than Dumping” by Eric Lipton on the “MediaClips” web site. At the end of the story, Lipton states that if people don’t recycle much,then it is too costly to have a recycling program, but if people recycle a lot, it becomeseconomically worthwhile. Model the situation as a strategic form game (say with just twopeople for simplicity) and show that there are two Nash equilibria: one in which both peoplerecycle a lot and one in which both people recycle little.The main point of the article is that New York’s recycling program is costly, and hence itis not cost-effective in the sense that the costs of the program are not offset by the moneygained by selling recycled materials. It would be cheaper to simply put everything intolandfills or incinerators. The costs are mainly collection costs: paying the sanitation workersto pick up the recyclables. However, near the end of the article, it is pointed out that ifcity residents recycled a significantly larger share of their trash, the program would begin tomake economic sense. So for simplicity say that there are two people, who can each recyclea little bit or a lot.

Recycle little Recycle lot

Recycle little 0, 0 0, -5

Recycle lot -5, 0 10, 10

The idea here is that if everyone recycles a lot, the city keeps its recycling program (sinceit is cost-effective) and everyone is happy. However, if everyone recycles a little bit, thenthe city loses its recycling program, which is not as nice. If only one person recycles a lot,then the recycling program is not profitable and hence there is no recycling program, andthe person who recycles a lot (since he is the only one) has to deliver her recyclables herselfto the recycling station, which is very inconvenient. Here there are two pure strategy Nashequilibria: (Recycle little, Recycle little) and (Recycle lot, Recycle lot).

21. Read “Drifter Jailed on Girls’ Lies Set Course of Desperation” by H. G. Reza, ChristineHanley, and James Ricci on the “Media Clips” web site. Model the situation as a strategicform game. Is this a Prisoners’ Dilemma? Why is it standard procedure for police tointerview witnesses separately?We can simplify the situation to include just two girls. I would say that each girl can chooseto lie or tell the truth, and the game looks like this:

Lie Truth

Lie 5, 5 -10, 0

Truth 0,-10 1, 1

If both girls lie, then they have an excuse for getting home late from school, which is betterfor them than if they both tell the truth. If one girl lies and the other tells the truth, theliar gets into big trouble; the one telling the truth does not get into trouble but feels bad forthe other girl.Here each girl wants to lie only if the other lies also. This not a Prisoners’ Dilemma, sinceno strategy is strongly dominated. There are two pure strategy Nash equilibria: (Lie, Lie)and (Truth, Truth).I think of the situation this way. It is “safer” to tell the truth, since you avoid gettinginto trouble (the -10 payoff). If both girls lie, the situation is better for both girls and issustainable, but before one girl lies, she would have to be quite sure that the other will lie

also. By allowing the girls to talk to each other, the girls could assure each other than theywould all lie (and also have the same story).

22. Read the excerpt from Richard Wright’s Black Boy on the “Media Clips” web site (downat the bottom of the page under “Text documents”). Model this as a strategic form gameand interpret the situation and the outcome in terms of the game and your predictions giventhe game.I would model it as a coordination game, something like this:

Harrison fights Harrison pretends

Wright fights -10, -10 0, -100

Wright pretends -100, 0 5, 5

Here both intend to pretend to fight and earn a few dollars. However, if you pretend to fightand the other person fights for real, then you get a very bad payoff. Hence if you have theslightest doubt that the other person is not pretending, you fight also. Hence you both endup fighting, which is much worse than if you both kept pretending.

23. Read “9 Questions about Syria You Were Too Embarrassed to Ask” by Max Fisher atthe “Media Clips” link. On page 6, under the “You didn’t answer my question” heading, asituation concerning chemical weapons is described. Model this as a strategic form game.According to the article, if a country believes its adversary is going to use chemical weapons(CW), it has a strong incentive to use them also. A country will not use chemical weaponsif it is certain that its opponent will not use them. Both sides are better off if neither useschemical weapons. So if we have two countries, we have something like this:

CW not

CW -10, -10 -5, -50

not -50, -5 0, 0

In this game, there are two (pure strategy) NE: (CW, CW) and (not, not).Say that country 1 now wants to use chemical weapons. The game now looks like

CW not

CW -10, -10 5, -50

not -50, -5 0, 0

Now the only (pure strategy) NE is (CW, CW). Country 1 wanting to use chemical weaponsmakes the situation devolve into one in which both countries use chemical weapons. Thenorm against chemical weapons is now broken.Say a superpower punishes any country which uses chemical weapons: whenever a countryuses chemical weapons, they get some inbound cruise missiles which adds an additional -20to their payoff. Then the game looks like this:

CW not

CW -30, -30 -15, -50

not -50, -25 0, 0

Now we have two (pure strategy) NE again: (CW, CW) and (not, not). Now the normagainst chemical weapons is restored because of the superpower’s threat.

24. Find all Nash (mixed strategy and pure strategy) equilibria to this version of the“Chicken” game:

[q] [1− q]

2 swerves 2 doesn’t

[p] 1 swerves 1, 1 0, 5

[1− p] 1 doesn’t 5, 0 -10, -10

ChickenWe can see that (swerves, doesn’t) and (doesn’t, swerves) are pure strategy Nash equilibria.To find the mixed strategy Nash equilibria, let p be the probability player 1 swerves and qbe the probability player 2 swerves.If player 2 swerves with probability q, then player 1’s expected payoff from swerving is(1)(q) + (0)(1− q) = q and her expected payoff from not swerving is (5)(q) + (−10)(1− q) =−10 + 15q. Now if q = 1, we can substitute to find that 1’s expected payoff from swervingis 1 and from not swerving is 5. So player 1’s best response is to set p = 0, that is, tonot swerve for sure. Now if q = 0, we can substitute to find that 1’s expected payoff fromswerving is 0 and from not swerving is -10. So player 1’s best response is to set p = 1, thatis, to swerve for sure. We can do this for various values of q and find player 1’s best responsecurve. When q is near 1, player 1’s best response is to set p = 0. When q is near 0, player1’s best response is to set p = 1. At some value of q, player 1 will “switch over” from playingp = 0 to p = 1. If we set the expected value of swerving equal to the expected value of notswerving, we get q = −10 + 15q. We solve this to find q = 10/14 = 5/7. At q = 5/7, player1’s expected value for swerving and not swerving are the same. Therefore, any p between 0and 1 will be a best response (in fact he will get the same expected utility no matter whatvalue of p he chooses). So we can draw player 1’s best response curve. We can similarlyfind player 2’s best response curve (check what happens at the extremes when p = 0 andp = 1, and calculate the “switchover” value of p which makes player 2’s expected value fromswerving and not swerving the same) to get the best response diagram.

p = 1

p = 0

q = 1 q = 0

Player 2’s bestresponse curve

(dotted line)


(solid line)

‘‘switchover’’ value p = 5/7

‘‘switchover’’ value q = 5/7

‘‘Chicken’’ game

Nash equilibrium: player 1doesn’t swerve with

probability 1, player 2swerves with probability 1

Nash equilibrium:player 1 swerves withprobability 1, player 2doesn’t swerve with

probability 1

Nash equilibrium: player 1 swerves withprobability 5/7 and doesn’t swerve with

probability 2/7, player 2 swerves withprobability 5/7 and doesn’t swerve with

probability 2/7

So the three Nash equilibria are where these two graphs intersect. There are the two purestrategy Nash equilibria we found earlier, and a mixed strategy Nash equilibrium in whichplayer 1 swerves with probability 5/7 and doesn’t swerve with probability 2/7 and player 2swerves with probability 5/7 and doesn’t swerve with probability 2/7.

25. Find all Nash equilibria to the “Early-late” game, which looks like this:

[q] [1− q]

2 arrives early 2 arrives late

[p] 1 arrives early 1, 1 -5, -1

[1− p] 1 arrives late -1, 0 3, 3

Early-lateWe do the same thing as before: find the value of q which makes player 1 indifferent betweenbeing early and late: we set (1)q + (−5)(1 − q) = (−1)(q) + 3(1 − q) and find that q =4/5. We find the value of p which makes player 2 indifferent between being early and late:(1)(p) + (0)(1− p) = (−1)(p) + (3)(1− p) and find that p = 3/5. We draw the best responsecurves (the diagram follows) and find that there are three Nash equilibria: two pure ones,(early, early) and (late, late), and one mixed one in which player 1 is early with probability3/5 and late with probability 2/5 and player 2 is early with probability 4/5 and late withprobability 1/5.

p = 1

p = 0

q = 1 q = 0


(dotted line)


(solid line)

‘‘switchover’’ value p = 3/5

‘‘switchover’’ value q = 4/5

‘‘Early-late’’ game

Nash equilibrium: player 1 isearly with probability 1,

player 2 is early withprobability 1

Nash equilibrium: player1 is late with probability1, player 2 is late with

probability 1

Nash equilibrium: player 1 is late withprobability 1/2 and isn’t late with

probability 1/2, player 2 is late withprobability 4/5 and isn’t late with

probability 1/5

26. Say you have an admirer whom you don’t like very much. You can either go to the libraryor the coffee shop to study. You prefer the coffee shop but you want to avoid your admirer.Your admirer can also go to the library or coffee shop to study. Your admirer prefers thelibrary but wants to be where you are more than anything else. So the game looks like:

Admirer goes to library Admirer goes to coffeeshop

You go to library 0, 3 4, 0

You go to coffeeshop 6, 0 0, 1

a. Find all (pure strategy and mixed strategy) Nash equilibria of this game.It is easy to verify that there are no pure strategy Nash equilibria of this game. To findmixed strategy Nash equilibria, let p be the probability you go to the library and let q be theprobability that your admirer goes to the library. To find out your best response curve, weneed to find the “switchover” value of q, that is, the q which makes you indifferent betweengoing to the library and going to the coffeeshop. So (0)(q) + (4)(1− q) = (6)(q) + (0)(1− q).Hence 4 = 10q and so q = 2/5. To find out your admirer’s best response curve, we need tofind the “switchover” value of p, that is, the p which makes your admirer indifferent betweengoing to the library and going to the coffeeshop. So (3)(p) + (0)(1− p) = (0)(p) + (1)(1− p).Hence 4p = 1 and so p = 1/4. The only Nash equilibrium is where you go to the librarywith probability 1/4 and the coffeeshop with probability 3/4, and your admirer goes to thelibrary with probability 2/5 and the coffeeshop with probability 3/5.

b. Now say that you begin to actually enjoy your admirer’s company. The game is now:Admirer goes to library Admirer goes to coffeeshop

You go to library 4, 3 0, 0

You go to coffeeshop 0, 0 6, 1

Find all (pure strategy and mixed strategy) Nash equilibria of this game.It is easy to verify that (library, library) and (coffeeshop, coffeeshop) are the only pure strat-egy Nash equilibria. To find mixed strategy Nash equilibria, again let p be the probabilityyou go to the library and let q be the probability that your admirer goes to the library. The“switchover” value of q is determined by the equation (4)(q)+(0)(1−q) = (0)(q)+(6)(1−q),so 4q = 6 − 6q, so 10q = 6, and hence q = 3/5. The “switchover” value of p is determinedby the equation (3)(p) + (0)(1− p) = (0)(p) + (1)(1− p), so 3p = 1− p, and hence 4p = 1,so p = 1/4. (Note that since your admirer’s payoffs did not change, this is just the same asbefore.) So the mixed strategy Nash equilibrium is where you go the library with probabil-ity 1/4 and the coffeeshop with probability 3/4, and your admirer goes to the library withprobability 3/5 and the coffeeshop with probability 2/5.

27. [from Spring 2002 midterm] Person 1 and Person 2 are competing for the affections ofthe extremely attractive Sandy. Person 1 and Person 2 plan to show up at Sandy’s houseat the same time on Saturday night to ask Sandy out. Since this is southern California, acrucial decision for both Person 1 and Person 2 is what kind of car to drive to Sandy’s house.Person 1 is wealthy and can either have the butler prepare the impressive Lamborghini orthe cute VW. Person 2 ordinarily drives a pickup truck, but can borrow a Lexus for the nightfrom a roommate. If Sandy sees the Lamborghini and the Lexus pull up, Sandy chooses theLamborghini because it is of course more impressive. If Sandy sees the Lamborghini andthe pickup truck, Sandy chooses the pickup truck because the Lamborghini is clearly tryingtoo hard. If Sandy sees the VW and the Lexus, Sandy chooses the Lexus because it is moreimpressive than the VW. If Sandy sees the VW and the pickup truck, Sandy chooses theVW because it is obviously more comfortable than the pickup truck.

a. Model this as a strategic form game between Person 1 and Person 2 (assume Sandy is nota player).Say that taking out Sandy yields a payoff of 1 and not taking Sandy out yields a payoff of0. We then get the following game.

Pickup truck Lexus

Lamborghini 0,1 1,0

VW 1,0 0,1

b. Find all (pure strategy and mixed strategy) Nash equilibria of this game.It is easy to see that there are no pure strategy Nash equilibria of this game. For mixedstrategy Nash equilibria, say that person 1 plays Lamborghini with probability p and VWwith probability 1 − p. Say that person 2 plays Pickup truck with probability q and Lexuswith probability 1− q.If person 1 plays Lamborghini, her expected utility is 0q+ 1(1− q) = 1− q. If person 1 playsVW, her expected utility is 1q + 0(1 − q) = q. To find the “switchover probability,” we setthese equal: 1− q = q and hence q = 1/2.If person 2 plays Pickup truck, his expected utility is p + 0(1 − p) = p. If person 2 playsLexus, his expected utility is 0p+ 1(1− p) = 1− p. To find the “switchover probability,” weset these equal: p = 1− p and so p = 1/2.Thus in the mixed Nash equilibrium, person 1 plays Lamborghini with probability 1/2 andVW with probability 1/2; person 2 plays Pickup truck with probability 1/2 and Lexus withprobability 1/2.

28. [from Spring 2002 midterm] Consider the following game.

2a 2b 2c

1a 0, 9 3, 0 1, 5

1b 1, 2 5, 4 4, 3

1c 0, 6 2, 1 6, 7

a. Find all pure strategy Nash equilibria of this game.Using our ∗ and + notation to indicate best responses, we get

2a 2b 2c

1a 0, 9+ 3, 0 1, 5

1b ∗1, 2 ∗5, 4+ 4, 3

1c 0, 6 2, 1 ∗6, 7+

So (1b, 2b) and (1c, 2c) are the pure strategy Nash equilibria.

b. Use the method of iterative elimination of (strongly or weakly) dominated strategies toeliminate as many strategies as possible (i.e. keep on eliminating until you can’t eliminateany more).It’s easy to see that 1b strongly dominates 1a and hence person 1 will never play 1a. Next,we iteratively eliminate 2a (once 1a is eliminated, 2c strongly dominates 2a). We thus havethe following game left over.

2b 2c

1b 5, 4 4, 3

1c 2, 1 6, 7

c. Find all mixed strategy Nash equilibria of this game. (Note: an answer like “p = 2/3, q =2/5” is not sufficient. Please write down in a sentence which strategies are played with whatprobability.)Say that person 1 plays 1b with probability p and 1c with probability 1− p. Say that person2 plays 2b with probability q and 2c with probability 1− q.If person 1 plays 1b, her expected utility is 5q + 4(1− q). If person 1 plays 1c, her expectedutility is 2q+6(1−q). To find the “switchover probability,” we set these equal: 5q+4(1−q) =2q + 6(1− q), or in other words 4 + q = 6− 4q. Hence 5q = 2 and so q = 2/5.If person 2 plays 2b, his expected utility is 4p + 1(1− p). If person 2 plays 2c, his expectedutility is 3p+7(1−p). To find the “switchover probability,” we set these equal: 4p+1(1−p) =3p + 7(1− p), or in other words 1 + 3p = 7− 4p. Hence 7p = 6 and so p = 6/7.Thus in the mixed Nash equilibrium, person 1 plays 1b with probability 6/7 and 1c withprobability 1/7; person 2 plays 2b with probability 2/5 and 2c with probability 3/5.

29. [from Spring 2002 final] In a simplified version of “Battleship,” say that there are fourspaces, numbered 1, 2, 3, 4. Person 1 chooses to fire a missle at one of these four spaces.Person 2 has a ship which is two spaces long, and chooses where to put the ship on theboard: she can either put it on spaces 1 and 2, on spaces 2 and 3, or on spaces 3 and 4. Thepeople make their choices simultaneously.

a. Model this as a strategic form game and find all mixed-strategy and pure-strategy Nashequilibria.Person 1’s possible strategies are 1, 2, 3, 4 (which space to fire the missile at). Person 2’spossible strategies are 12, 23, 34 (put the ship on spaces 1 and 2, put the ship on spaces 2and 3, or put the ship on spaces 3 and 4). Say that you get a payoff of 1 for winning and 0for losing. We thus get the following game.

12 23 34

1 1, 0 0, 1 0, 1

2 1, 0 1, 0 0, 1

3 0, 1 1, 0 1, 0

4 0, 1 0, 1 1, 0

It’s easy to see that there are no pure strategy Nash equilibria. To find mixed strategy Nashequilibria, we can simplify the game first. For person 1, firing at 2 weakly dominates firingat 1 and firing at 3 weakly dominates firing at 4. So we can eliminate firing at 1 and firingat 4 and get the following.

12 23 34

2 1, 0 1, 0 0, 1

3 0, 1 1, 0 1, 0

Now it is easy to see that for person 2, putting the ship at 23 is weakly dominated by puttingit at 34 or 12. So we eliminate 23 and get the following game.

12 34

2 1, 0 0, 1

3 0, 1 1, 0

To find the mixed strategy Nash equilibria, say that person 1 fires at 2 with probability pand fires at 3 with probability 1−p. Say that person 2 places the boat at 12 with probabilityq and places the boat at 34 with probability 1− q.If person 1 fires at 2, her expected utility is 1q + 0(1 − q) = q. If person 1 fires at 3, herexpected utility is 0q + 1(1− q) = 1− q. To find the “switchover probability,” we set theseequal: q = 1− q and hence q = 1/2.If person 2 places the boat at 12, his expected utility is 0p + 1(1 − p) = 1 − p. If person2 places the boat at 34, his expected utility is 1p + 0(1 − p) = p. To find the “switchoverprobability,” we set these equal: 1− p = p and so p = 1/2.Thus in the mixed Nash equilibrium of the original game, person 1 fires at 2 with probability1/2 and fires at 3 with probability 1/2; person 2 places the boat at 12 with probability 1/2and places the boat at 34 with probability 1/2.

b. Now say that there are 5 spaces, numbered 1, 2, 3, 4, 5. Model this as a strategic formgame and find all mixed-strategy and pure-strategy Nash equilibria.Similarly, we get the following game.

12 23 34 45

1 1, 0 0, 1 0, 1 0, 1

2 1, 0 1, 0 0, 1 0, 1

3 0, 1 1, 0 1, 0 0, 1

4 0, 1 0, 1 1, 0 1, 0

5 0, 1 0, 1 0, 1 1, 0

Again, there are no pure Nash equilibria of this game. As before, for person 1, firing at 1 isweakly dominated by firing at 2, and firing at 5 is weakly dominated by firing at 5. So weget the following.

12 23 34 45

2 1, 0 1, 0 0, 1 0, 1

3 0, 1 1, 0 1, 0 0, 1

4 0, 1 0, 1 1, 0 1, 0

Now for person 2, placing the boat at 23 is weakly dominated by placing the boat at 12,and placing the boat at 34 is weakly dominated by placing the boat at 45. So we get thefollowing.

12 45

2 1, 0 0, 1

3 0, 1 0, 1

4 0, 1 1, 0

Now for person 1, firing at 3 is weakly dominated by firing at 2 (or at 4). Hence we havethe following.

12 45

2 1, 0 0, 1

4 0, 1 1, 0

We can find the mixed strategy Nash equilibrium as before. We find that in the mixed Nashequilibrium of this game (and hence the original game), person 1 fires at 2 with probability1/2 and fires at 4 with probability 1/2; person 2 places the boat at 12 with probability 1/2and places the boat at 45 with probability 1/2.

30. Say that a seller tries to sell a car to a buyer. The car is worth $1000 to the seller and$2000 to the buyer, and both people know this. First, the seller proposes a price of either$1800 or $1200 to the buyer. Given this price, the buyer can either accept or reject the offer.Model this as an extensive form game.

a. Find all Nash equilibria.The easiest way to find all Nash equilibria of this game is to represent it as a strategic formgame:

if $1800, accept; if $1800, accept; if $1800, reject; if $1800, reject;

if $1200, accept if $1200, reject if $1200, accept if $1200, reject

$1800 800, 200 800, 200 0, 0 0, 0

$1200 200, 800 0, 0 200, 800 0, 0

where the seller chooses the row and the buyer chooses the column. Then it is clear that thethree (pure strategy) Nash equilibria are ($1800, if $1800, accept; if $1200, accept), ($1800,if $1800, accept; if $1200, reject), and ($1200, if $1800, reject; if $1200, accept). Note thatthere is a Nash equilibrium where the buyer “holds out” for a price of $1200 and succeeds.

b. Which Nash equilibria are subgame perfect?We note that in a subgame perfect Nash equilibrium, the buyer will accept either offer. Sothe only subgame perfect Nash equilibrium is ($1800, if $1800, accept; if $1200, accept).If you want to do this with a game tree, the game in extensive form looks like this:

Seller

Buyer

Buyer

offer $1800

offer $1200

accept

accept

reject

reject

800, 200

200, 800

0, 0

0, 0

By backward induction, we find the following subgame Nash equilibrium:

Seller

Buyer

Buyer

offer $1800

offer $1200

accept

accept

reject

reject

800, 200

200, 800

0, 0

0, 0

31. Say that you and your brother divide up 4 oranges in the following manner. First, youdivide the oranges up into two piles: the left pile and the right pile. For example, you couldput one orange in the left pile and three in the right pile. Then, your brother decides whichof the two piles he wants; he keeps that pile and you get the other pile. Both of you likeoranges.

a. Model this as an extensive form game and find the subgame perfect Nash equilibria. Howwill the oranges be divided?You have five possible strategies: put 4 oranges in the left pile and 0 in the right, 3 in theleft and 1 in the right, 2 in the left and 2 in the right, 1 in the left and 3 in the right, and 0in the left and 4 in the right (I abbreviate these as 4L, 0R, etc.) Your brother can choose ineach of these contingencies whether to choose the left or right pile. We can say that payoffsare just the number of oranges one gets. Hence the extensive form game looks like this:

You

L

R

L

R

L

R

L

R

L

R

0, 4

4L, 0R

3L, 1R

1L, 3R

0L, 4R

2L, 2R

Bro

Bro

Bro

Bro

Bro0, 4

4, 0

4, 0

1, 3

1, 3

3, 1

3, 1

2, 2

2, 2

There are two subgame perfect Nash equilibria. The first, written as arrows in a game tree,is:

You

L

R

L

R

L

R

L

R

L

R

0, 4

4L, 0R

3L, 1R

1L, 3R

0L, 4R

2L, 2R

Bro

Bro

Bro

Bro

Bro

0, 4

4, 0

4, 0

1, 3

1, 3

3, 1

3, 1

2, 2

2, 2

The second, written as arrows in a game tree, is:

You

L

R

L

R

L

R

L

R

L

R

0, 4

4L, 0R

3L, 1R

1L, 3R

0L, 4R

2L, 2R

Bro

Bro

Bro

Bro

Bro

0, 4

4, 0

4, 0

1, 3

1, 3

3, 1

3, 1

2, 2

2, 2

If we were to spell these out explicitly, in the first equilibrium, person 1 plays strategy “2L,2R” and person 2 plays strategy “If 4L, 0R, choose L; if 3L, 1R, choose L; if 2L, 2R, chooseL; if 1L, 3R, choose R; if 0L, 4R, choose R”. The second equilibrium is similar. In both ofthese equilibria, you and your brother split the 4 oranges equally. Note that your brother’sstrategy is the choice of either the left or right pile conditional on your division.

32. Say that country A can either attack or not attack country B. Country B cannot defenditself; the only thing it can do is to set off a “doomsday device” which obliterates bothcountries. The worst thing for both countries is to be obliterated; for country A, the bestthing is to attack and not be obliterated; for country B, the best thing is to not be attacked.

a. Think of country A moving first and model this as an extensive form game. Find all purestrategy Nash equilibria and find the subgame perfect Nash equilibria.The game in extensive form looks like this (of course, your choice of payoffs might be differentfrom mine):

A

B

B

-10, -10

-10, -10

0, 0

5, -5attack

don’t

obliterate

don’t

obliterate

don’t

The strategic form looks like this:if attack, obliterate; if attack, obliterate; if attack, don’t; if attack, don’t;

if not, obliterate if not, don’t if not, obliterate if not, don’t

attack -10,-10 -10,-10 5,-5 5,-5

don’t -10,-10 0,0 -10,-10 0, 0

There are three pure strategy Nash equilibria: (don’t, if attack, obliterate; if not, don’t),(attack, if attack, don’t; if not, obliterate), (attack, if attack, don’t; if not, don’t). Bybackwards induction or whatever other method, the subgame perfect Nash equilibrium is(attack, if attack, don’t; if not, don’t).

b. Now think of country B moving first and model this as an extensive form game. Find allpure strategy Nash equilibria and find the subgame perfect Nash equilibria.The game in extensive form looks like this (even though now B moves first, I write countryA’s payoffs first and country B’s payoffs second, so to remain consistent with part a. above).

B

A

A

-10, -10

-10, -10

5, -5

0, 0

obliterate

don’t

attack

don’t

attack

don’t

The strategic form looks like this:obliterate don’t

if obliterated, attack; if not, attack -10,-10 5,-5

if obliterated, attack; if not, don’t -10,-10 0,0

if obliterated, don’t; if not, attack -10,-10 5,-5

if obliterated, don’t; if not, don’t -10,-10 0,0

The two Nash equilibria are (if obliterated, attack; if not, attack, don’t) and (if obliterated,don’t; if not, attack, don’t). These are both subgame perfect Nash equilibria.

33. The game called “Nim” goes like this: there are a pile of five stones on the ground.Player 1 can take either 1 or 2 stones. Then player 2 can take either 1 or 2 stones. Theycontinue taking either 1 or 2 stones in turn until all the stones are gone. The player whotakes the last stone (or stones) wins.

a. Model this as an extensive form game and find the subgame perfect Nash equilibria.The extensive form game is shown below.

player 1

player 2

player 2

take 1 stone

take 2 stones

player 1

player 1

take 1 stone

take 2 stones

player 2

1, -1

take 1 stone

1, -1take 2 stones

take 1 stone

-1, 1

take 1 stone

player 1

take 2 stones player 2

take 1 stone

1, -1

take 2 stones

-1, 1

take 1 stone

player 1

take 1 stone

player 2

take 1 stone

player 1

take 1 stone

1, -1

take 1 stone

-1, 1

take 2 stones

player 2

take 2 stones-1, 1

take 1 stone*

*

*

Arrows in the game tree show one of the subgame perfect Nash equilibria. There are actuallyfour subgame perfect Nash equilibria, because there are ties at the starred nodes. I won’tactually spell out all of the four subgame perfect Nash equilibria, since it would be tedious.In all of the subgame perfect Nash equilibria, player 1 initially takes 2 stones, and then nomatter what player 2 then does, player 1 will eventually win.

b. What happens when you start with n stones instead of five?You can show that if n is a multiple of 3, then in any subgame perfect Nash equilibrium,player 2 wins. Player 2 plays the following strategy: if player 1 picks one stone, then player2 picks two stones; if player 1 picks two stones, then player 2 picks one stone. This wayplayer 2 ensures that the remaining stones left after she plays is always a multiple of three.

If n is not a multiple of 3, then in any subgame perfect Nash equilibrium, player 1 wins.Player 1 initially picks up whatever number of stones which makes the remaining numberof stones a multiple of three. From then on, if player 2 picks two stones, player 1 picks onestone, and if player 2 picks one stone, player 1 picks two stones.

34. Say we have a version of the “centipede” game:

1 2 1 2 8, 8a a a a

2, 2 1, 8 4, 4 2, 16

d d d d

a. Find the subgame perfect Nash equilibrium.We can write down the subgame perfect equilibrium in the game tree below:

1 2 1 2 8, 8a a a a

2, 2 1, 8 4, 4 2, 16

d d d d

At each “node,” each player plays down. Person 1’s strategy is “First play down; if person2 played across, play down.” Person 2’s strategy is “If person 1 played across, play down; ifperson 1 plays across again, then play down.”

b. Represent this game in strategic (matrix) form.Remember that a strategy is a complete contingent plan. Person 1 can move either across ordown, and in the contingency that person 2 moves across, can move either across or down.So we will represent his strategies as aa, ad, da, and dd. If Person 1 moves across, Person 2can move either across or down, and in the contingency that person 1 moves across again,can move either across or down. So we will represent his strategies also as aa, ad, da, anddd. So the strategic form game looks like this:

aa ad da dd

aa 8, 8 2, 16 1, 8 1, 8

ad 4, 4 4, 4 1, 8 1, 8

da 2, 2 2, 2 2, 2 2, 2

dd 2, 2 2, 2 2, 2 2, 2

c. Solve this strategic form game using iterated elimination of weakly dominated strategies.Show the order of elimination.By inspection, we can see that for person 2, dd and ad weakly dominate aa. So we are leftwith

ad da dd

aa 2, 16 1, 8 1, 8

ad 4, 4 1, 8 1, 8

da 2, 2 2, 2 2, 2

dd 2, 2 2, 2 2, 2

Now for person 1, both da and dd weakly dominate aa. So we are left withad da dd

ad 4, 4 1, 8 1, 8

da 2, 2 2, 2 2, 2

dd 2, 2 2, 2 2, 2

Now it is clear that for person 2, both da and dd weakly dominate ad. Hence we are leftwith

da dd

ad 1, 8 1, 8

da 2, 2 2, 2

dd 2, 2 2, 2

Now for person 1, both da and dd weakly dominate ad. So finally, we haveda dd

da 2, 2 2, 2

dd 2, 2 2, 2

It is clear that there can be no more domination. So the prediction is that person 1 will playeither da or dd, and person 2 will play either da or dd. In any of these cases, the outcomewill be that they both get a payoff of 2.

35. [from Spring 2004 midterm] Boyfriend and Girlfriend are on a 7 day cruise, which startson Monday and goes until Sunday. Both know that this cruise is their last hurrah. Bothknow that once Sunday comes around and the cruise ends, if they are still together, the gameends because Boyfriend will be forced to dump Girlfriend (because otherwise he would bedisowned by his wealthy parents).On Monday evening, Boyfriend decides whether to break up or continue in the relationship.If Boyfriend decides to continue, then on Tuesday evening, Girlfriend decides whether tobreak up or to continue. If Girlfriend decides to continue, then on Wednesday evening,Boyfriend decides whether to continue. The game continues like this, with Boyfriend andGirlfriend taking turns deciding whether to continue or to break up (Boyfriend gets todecide on Monday, Wednesday, and Friday, and Girlfriend decides on Tuesday, Thursday,and Saturday). If anyone decides to break up, the game is immediately over. If no onedecides to break up, then they make it to Sunday, and Girlfriend is left heartbroken.Each person enjoys the others company, and gets a payoff of 5 for every day the relationshipcontinues. However, no person wants to be dumped; if a person breaks up the relationship,the other person (the dumped one) gets 10 added to her payoff. For example, if they makeit all the way to Sunday, Boyfriend’s payoff is 35 and Girlfriend’s payoff is 25.

a. Represent this as an extensive form game.

It’s just like the centipede game above. Note that on Sunday, Boyfriend has only one move.One could have written it so that Girlfriend has the last move, on Saturday.

break

cont

5, -10

Boy

break

cont

0, 10

Girl

break

cont

15, 5

Boy

break

cont

10, 20

Girl

break

cont

25, 15

Boy

break

cont

20, 30

Girl

break

35, 25

Boy

(Monday) (Tuesday) (Wednesday) (Thursday) (Friday) (Saturday) (Sunday)

b. Find the subgame perfect Nash equilibrium.The subgame perfect Nash equilibrium is shown by the arrows above. Even though theyenjoy each other’s company, Boyfriend and Girlfriend want to dump before being dumped.Hence Boyfriend dumps immediately, which is a sad thing for all.

36. Consider the “cross-out game.” In this game, one writes down the numbers 1, 2, 3.Person 1 starts by crossing out any one number or any two adjacent numbers: for example,person 1 might cross out 1, might cross out 1 and 2, or might cross out 2 and 3. Then person2 also crosses out either one number or two adjacent numbers. For example, starting from1, 2, 3, say person 1 crosses out 1. Then person 2 can either cross out 2, cross out 3, orcross out both 2 and 3. Play continues like this. Once a number is crossed out, it cannotbe crossed out again. Also, if for example person 1 crosses out 2 in her first move, person2 cannot then cross out both 1 and 3, because 1 and 3 are not adjacent (even though 2 iscrossed out). The winner is the person who crosses out the last number.

a. Model this as an extensive form game. Show a subgame perfect Nash equilibrium of thisgame by drawing appropriate arrows in the game tree.We did this in class.

b. Now instead of just three numbers, say that you start with m numbers. In other words,you have the numbers 1, 2, 3, . . . ,m. Can person 1 always win this game? (Hint: look atm = 4, m = 5, etc. first to get some ideas.)Person 1 can always win this game. His strategy is follows. If m is even, then person 1 startsby crossing out the two middle numbers. If m is odd, then person 1 starts out by crossing outthe middle number. Henceforth, whatever number(s) person 2 crosses out, person 1 respondsby crossing out the “mirror image” of those number(s) on the “other side” (if person 2 crossesout the number x, person 1 crosses out m− x + 1). Play continues like this, with each timeperson 1 responding in this manner to person 2. It is easy to see that person 1 is guaranteedto cross out the last number.For example, say that m = 7. Person 1 crosses out 4, the middle number. If person 2 crossesout 1, then person 1 responds by crossing out 7. If person 2 then crosses out 5 and 6, thenperson 1 responds by crossing out 2 and 3.Notice that this game is not related to the Nim game in the homework (there is no issue ofmultiples of 3, for example). Answers of the form “Person 1 will always win by crossing outthe middle numbers” receive no credit without a full explanation like the one above.

37. Read “It’s Not Just About Bonds, It’s About Who’s on Deck” by Jack Curry on the“Media Clips” website. Model the situation using extensive form games and use the gamesto show that who bats behind Barry Bonds influences a pitcher’s decision of how to pitch toBonds.I model the situation in this way.

Giants

Dodgers

Dodgers

Cruz

Alfonzo

Walk Bonds

Don’t walk Bonds

Don’t walk Bonds

Walk Bonds

0.30, 0.70

0.40, 0.60

0.50, 0.50

0.40, 0.60

Here the Giants can either put in Alfonzo or Cruz behind Bonds. The Dodgers can eitherintentionally walk Bonds or not. If Alfonzo bats behind Bonds, and the Dodgers intentionallywalk Bonds, putting him on base, they run the risk that Alfonzo will advance Bonds. ButAlfonzo is hitting poorly enough these days that there is only a 30 percent chance that Bondswill advance. If the Dodgers do not intentionally walk Bonds and pitch to him, then Bondswill get on base with a 40 percent chance. If the Giants put in Cruz behind Bonds, then ifthe Dodgers intentionally walk Bonds, they would have to face Cruz, who is on a hot streakand will advance Bonds with a 50 percent chance. As before, if the Dodgers pitch to Bonds,then Bonds will get on base with a 40 percent chance.The subgame perfect Nash equilibrium is shown in the game tree also. If the Giants putAlfonzo behind Bonds, then the Dodgers will walk Bonds, knowing that Alfonzo is slumping.If the Giants put Cruz behind Bonds, then the Dodgers will pitch to Bonds. Hence the Giantsput Cruz behind Bonds.

38. Richard Nixon subscribed to a foreign-policy principle which he called the “madmantheory” or the “theory of excessive force” (see “The Trials of Henry Kissinger” on the MediaClips website, or read “Nixon’s Nuclear Ploy” by William Burr and Jeffrey Kimball, also onthe Media Clips website). We will explain this theory using an extensive form game.Say that the US is facing an adversary (for example North Vietnam). The US makes thefirst move: it can either threaten North Vietnam or do nothing. If the US does nothing, thenthe game ends and “nothing happens.” If the US makes a threat, North Vietnam can eithercomply or not comply. If North Vietnam complies, then the game ends. If North Vietnamdoes not comply, then the US can either strike militarily or not.

a. Say the payoffs are like this: if the US does nothing, then both get payoff 0. If the USthreatens and North Vietnam complies, then the US gets payoff 10 and North Vietnam getspayoff -5. If the US threatens, North Vietnam does not comply, and the US strikes, then theUS gets payoff -10 and North Vietnam gets payoff -10. If the US threatens, North Vietnamdoes not comply, and the US does not strike, then the US gets payoff -5 and North Vietnamgets payoff 10.The idea here is that the best thing for the US is for the US to make a threat and NorthVietnam to comply; this way the US doesn’t have to perform the military strike. The worstthing for the US is to have to perform the military strike, since this would risk expandingthe war. The best thing for North Vietnam is to openly defy the US and “call the bluff” ofthe US; the worst thing is the military strike. The second-worst thing is to comply to theUS threat, and the second-best thing is to not be threatened at all.Model this as an extensive form game and find the subgame perfect Nash equilibrium.The game looks like this.

US

US

NV

nothing

nothing

complythreaten

not comply

strike

0, 0

10, -5

-5, 10

-10, -10

The subgame perfect Nash equilibrium is shown with the arrows. Since a military strike iscostly to both sides, the US will not strike if North Vietnam does not comply. Hence NorthVietnam will not comply if threatened, and hence the US will not threaten. In a sense, ifthe US threatens, then North Vietnam will “call the bluff” because it knows that the US isnot prepared to face the losses of a military strike.

b. Now say that Nixon, by purposefully showing the world that he is a “madman,” convinceseveryone that the US payoff from a military strike is 5, not -10. In other words, Nixon tellsthe world that as far as he is concerned, he would enjoy a military strike. Model this asan extensive form game (all other payoffs are the same) and find the subgame perfect Nashequilibrium.Now the game looks like this.

US

US

NV

nothing

nothing

complythreaten

not comply

strike

0, 0

10, -5

-5, 10

5, -10

Now the US would love to engage in the military strike if North Vietnam does not comply.Hence North Vietnam complies, and hence the US can threaten.

c. Compare the two situations and explain why Nixon wants to show the world that he is a“madman.”Nixon is better off in the second scenario. If people believe that the second scenario is thecorrect one (that Nixon is a “madman” in the sense that he wants to attack even if highcasualties result) then Nixon can threaten and get his way without ever actually having toresort to military action.

39. Say we have two basketball players, Nash and Yao. Nash has the ball 20 feet from thebasket and Yao is defending. Nash can either take a jump shot or drive the basket. Yao caneither come out and contest the shot, or can stay in and protect the basket. The game lookslike this.

Yao comes out Yao protects

Nash takes jump shot 0.3, 0.7 0.4, 0.6

Nash drives 0.5, 0.5 0.3, 0.7

For example, if Nash takes a jump shot and Yao comes out, then Nash has a 30 percentchance of scoring and Yao has a 70 percent chance of successfully defending.

a. Find the mixed strategy Nash equilibrium of this game. In the mixed Nash equilibrium,what is Nash’s overall probability of scoring?It’s easy to see that there are no pure strategy Nash equilibria. To find mixed strategy Nashequilibria, let’s set up our p and q:

[q] [1− q]


[p] Nash takes jump shot 0.3, 0.7 0.4, 0.6

[1− p] Nash drives 0.5, 0.5 0.3, 0.7

First solve for q. If Nash takes a jump shot, he gets payoff 0.3(q)+0.4(1−q). If Nash drives,he gets payoff 0.5(q) + 0.3(1− q). To find the switchover value of q, we set these equal andget 0.3(q) + 0.4(1− q) = 0.5(q) + 0.3(1− q), or in other words 0.4− 0.1q = 0.3 + 0.2q. Hence0.1 = 0.3q and we have q = 1/3.

Now solve for p. If Yao comes out, he gets payoff 0.7p + 0.5(1 − p). If Yao protects,he gets 0.6p + 0.7(1 − p). To find the switchover value of p, we set these equal and get0.7p + 0.5(1 − p) = 0.6p + 0.7(1 − p), or in other words 0.5 + 0.2p = 0.7 − 0.1p. Hence0.3p = 0.2 and we have p = 2/3.So the mixed Nash equilibrium is that Nash takes a jump shot with probability 2/3 and driveswith probability 1/3 and Yao comes out with probability 1/3 and protects with probability2/3.If Nash takes a jump shot, his overall probability of scoring is 0.3(1/3) + 0.4(2/3) = 1.1/3 =0.37. If Nash drives, his overall probability of scoring is 0.5(1/3) + 0.3(2/3) = 1.1/3 = 0.37.Notice these numbers are the same because otherwise, Nash would not want to randomize.Hence Nash’s overall probability of scoring is 0.37.

b. Now say that Nash practices over the summer and improves his jump shot. Now the gamelooks like this.


Nash takes jump shot 0.4, 0.6 0.5, 0.5

Nash drives 0.5, 0.5 0.3, 0.7

Find the mixed strategy Nash equilibrium of this new game. In the mixed Nash equilibrium,what is Nash’s overall probability of scoring?Again, there are no pure strategy Nash equilibria. To find mixed strategy Nash equilibria,let’s set up our p and q:

[q] [1− q]


[p] Nash takes jump shot 0.4, 0.6 0.5, 0.5

[1− p] Nash drives 0.5, 0.5 0.3, 0.7

First solve for q. If Nash takes a jump shot, he gets payoff 0.4(q)+0.5(1−q). If Nash drives,he gets payoff 0.5(q) + 0.3(1− q). To find the switchover value of q, we set these equal andget 0.4(q) + 0.5(1− q) = 0.5(q) + 0.3(1− q), or in other words 0.5− 0.1q = 0.3 + 0.2q. Hence0.2 = 0.3q and we have q = 2/3.Now solve for p. If Yao comes out, he gets payoff 0.6p + 0.5(1 − p). If Yao protects,he gets 0.5p + 0.7(1 − p). To find the switchover value of p, we set these equal and get0.6p + 0.5(1 − p) = 0.5p + 0.7(1 − p), or in other words 0.5 + 0.1p = 0.7 − 0.2p. Hence0.3p = 0.2 and we have p = 2/3.So the mixed Nash equilibrium is that Nash takes a jump shot with probability 2/3 and driveswith probability 1/3 and Yao comes out with probability 2/3 and protects with probability1/3.If Nash takes a jump shot, his overall probability of scoring is 0.4(2/3) + 0.5(1/3) = 1.3/3 =0.43. If Nash drives, his overall probability of scoring is 0.5(2/3) + 0.3(1/3) = 1.3/3 = 0.43.Notice these numbers are the same because otherwise, Nash would not want to randomize.Hence Nash’s overall probability of scoring is 0.43.

c. After Nash improves his jump shot, does he go to his jump shot more often? After Nashimproves his jump shot, is it more successful? After Nash improves his jump shot, is his drivemore successful? Why does an improved jump shot improve other parts of Nash’s game?

When Nash improves his jump shot, he doesn’t go to it any more often; he still goes to it1/3 of the time. Nash’s jump shot improves from an overall 0.37 scoring probability to 0.43.Interestingly, his drive is more successful, also improving to an overall scoring probability of0.43. His drive is more successful even though he hasn’t practiced it (he only practiced hisjump shot). The reason his drive is more successful is because Yao has to come out moreoften (with probability 2/3 instead of 1/3) because of his improved jump shot. So it reallyis true that improving one aspect of your game improves the other aspects of your game,because it makes the defense react. Nash’s better jump shot makes Yao come out more,which improves the chances of Nash’s drive.

40. Consider the “Battle of the Sexes” game below.2a 2b

1a 2, 1 0, 0

1b 0, 0 1, 2

a. Find all Nash equilibria (pure strategy and mixed strategy) of this game.The pure Nash equilibria are (1a, 2a) and (1b, 2b). To find the mixed Nash equilibria, letperson 1 play 1a with probability p and 1b with probability 1− p. Let person 2 play 2a withprobability q and 2b with probability 1− q.Given person 2’s strategy, if person 1 plays 1a, he gets an expected payoff of 2q+0(1−q) = 2q.If person 1 plays 1b, he gets an expected payoff of 0q + 1(1− q) = 1− q. To find person 1’s“switchover” probability, we set 2q = 1− q and get q = 1/3.Given person 1’s strategy, if player 2 plays 2a, she gets an expected payoff of 1p+0(1−p) = p.If person 2 plays 2b, she gets an expected payoff of 0p + 2(1 − p) = 2 − 2p. To find person2’s “switchover” probability, we set p = 2− 2p and get p = 2/3.Thus in the mixed strategy Nash equilibrium, person 1 plays 1a with probability 2/3 andplays 1b with probability 1/3 and person 2 plays 2a with probability 1/3 and 2b with prob-ability 2/3.

b. Are any strategies in this game weakly or strongly dominated?No strategies are weakly or strongly dominated.

41. Consider the following game.2a 2b 2c 2d 2e

1a 63,−1 28,−1 −2, 0 −2, 45 −3, 19

1b 32, 1 2, 2 2, 5 33, 0 2, 3

1c 54, 1 95,−1 0, 2 4,−1 0, 4

1d 1,−33 −3, 43 −1, 39 1,−12 −1, 17

1e −22, 0 1,−13 −1, 88 −2,−57 −3, 72

a. Find all pure strategy Nash equilibria of this game.We can use our stars and pluses:

2a 2b 2c 2d 2e

1a ∗63,−1 28,−1 −2, 0 −2, 45+ −3, 19

1b 32, 1 2, 2 ∗2, 5+ ∗33, 0 ∗2, 31c 54, 1 ∗95,−1 0, 2+ 4,−1 0, 4

1d 1,−33 −3, 43+ −1, 39 1,−12 −1, 17

1e −22, 0 1,−13 −1, 88+ −2,−57 −3, 72

So the only pure strategy Nash equilibrium is (1b, 2c).

b. Make a prediction in this game by iteratively eliminating (strongly or weakly) dominatedstrategies.First we note that 1c strongly dominates 1d and 1e.

2a 2b 2c 2d 2e

1a 63,−1 28,−1 −2, 0 −2, 45 −3, 19

1b 32, 1 2, 2 2, 5 33, 0 2, 3

1c 54, 1 95,−1 0, 2 4,−1 0, 4

Next we note that 2c strongly dominates 2a and 2b.2c 2d 2e

1a −2, 0 −2, 45 −3, 19

1b 2, 5 33, 0 2, 3

1c 0, 2 4,−1 0, 4

Next we note that 1b strongly dominates 1a and 1c.2c 2d 2e

1b 2, 5 33, 0 2, 3

Finally, we note that 2c strongly dominates 2d and 2e.2c

1b 2, 5

So the prediction is that person 1 plays 1b and person 2 plays 2c.

42. [from Spring 2002 midterm] Mother can ask either Sister or Brother to do the dishes whileshe goes out shopping. If Mother asks Sister, she can either do it or not do it. If Mother asksBrother, he can either do it or not do it. Since Sister does a better job, Mother prefers Sisterdoing the dishes over Brother doing the dishes. However, Mother prefers Brother doing thedishes over them not being done.Both Sister’s and Brother’s preferences are like this: the best thing is for the other personto do the dishes; the second best thing is for Mother to ask the other person and have theother person not do it (since then the other person will get blamed). The third best thingis to do the dishes, and the worst thing is to be asked to do the dishes but then not do it(since you will get in trouble).The main thing here is to set up the players and strategies correctly. It is crucial that thisis represented as a 3 person game.

You do it You don’t

Mom asks you 5, 3, 9 0, 0, 7

Mom asks brother 4, 9, 3 4, 9, 3

Brother does it

You do it You don’t

Mom asks you 5, 3, 9 0, 0, 7

Mom asks brother 0, 7, 0 0, 7, 0

Brother doesn’t

a. Find all (pure strategy) Nash equilibria.By checking all the strategy profiles, we can see that (Mom asks you, You do it, Brotherdoes it), (Mom asks you, You do it, Brother doesn’t), and (Mom asks brother, You don’t,Brother does it) are the three Nash equilibria.

43. The simplest kind of game has two players, who each have two possible actions. We callthese games “2× 2 games.”

a. Write down a 2 × 2 game which has exactly one pure strategy Nash equilibrium and nomixed strategy Nash equilibrium. Solve for the equilibrium.A Prisoners’ Dilemma works here.

2a 2b

1a 3, 3 0, 4

1b 4, 0 1, 1

There is only one pure strategy Nash equilibrium (1b, 2b) and no mixed strategy Nash equi-librium (since person 1 will never play 1a and person 2 will never play 2a).

b. Write down a 2 × 2 game which has no pure strategy Nash equilibrium and exactly onemixed strategy Nash equilibrium. Solve for the equilibrium.The penalty kick game works here.

2a 2b

1a 1, 0 0, 1

1b 0, 1 1, 0

As we did in class, in the mixed strategy Nash equilibrium, person 1 plays 1a with probability1/2 and 1b with probability 1/2, and person 2 plays 2a with probability 1/2 and 2b withprobability 1/2.

c. Write down a 2× 2 game which has exactly three total (pure and mixed) Nash equilibria.Solve for the equilibria.The chicken game works here.

2a 2b

1a 3, 3 1, 4

1b 4, 1 0, 0

Here (1b, 2a) and (1a, 2b) are pure strategy Nash equilibria. If we compute the mixed strategyNash equilibrium, we find that person 1 plays 1a with probability 1/2 and 1b with probability1/2, and person 2 plays 2a with probability 1/2 and 2b with probability 1/2.

d. Write down a 2× 2 game in which the total number of (pure and mixed) Nash equilibriais neither one nor three. Solve for the equilibria.

The game in which “everything ties” works here.2a 2b

1a 3, 3 3, 3

1b 3, 3 3, 3

Here there are four pure strategy Nash equilibria: (1a, 2a), (1a, 2b), (1b, 2a), (1b, 2b). In fact,if person 1 plays any mixed strategy and person 2 plays any mixed strategy, then that is amixed strategy Nash equilibrium.

44. Say that country A and country I are at war. The two countries are separated by asystem of rivers, as shown below.

A b c

d e f

g h ICountry I sends a naval fleet with just enough supplies to reach A. The fleet must stop forthe night at intersections (for example, if the fleet takes the path IhebA, it must stop the firstnight at h, the second at e, and the third at b). If unhindered, on the fourth day the fleetwill reach A and destroy country A. Country A can send a fleet to prevent this. CountryA’s fleet has enough supplies to visit three contiguous intersections, starting from A (forexample Abcf). If it catches Country I’s fleet (that is, if both countries stop for the nightat the same intersection), it destroys the fleet and wins the war. Model this as a strategicform game, assuming that the winner gets payoff 1 and the loser gets payoff -1. Iterativelyeliminate weakly dominated strategies and make some sort of prediction.Please see the attached page.

45. [from Spring 2002 final] There are three people who each simultaneously choose the letterA or the letter B. If a person chooses a letter which no one else chooses, then he gets a payoffof 4; if a person chooses a letter which some other person chooses, then he gets a payoff of 0.The only exception is if everyone chooses the letter A, in which case everyone gets a payoffof 3.

a. Model this as a strategic form game and find all pure-strategy Nash equilibria.The game looks like this:

2A 2B

1A 3, 3, 3 0, 4, 0

1B 4, 0, 0 0, 0, 4

3A

2A 2B

1A 0, 0, 4 4, 0, 0

1B 0, 4, 0 0, 0, 0

3BThe pure strategy Nash equilibria are (1B, 2A, 3A), (1A, 2B, 3A), (1B, 2B, 3A), (1A, 2A, 3B),(1B, 2A, 3B), and (1A, 2B, 3B).

b. There exists a mixed-strategy Nash equilibrium in which each person plays A with prob-ability p and B with probability 1− p. Find p.Say person 2 plays 2A with probability p and 2B with probability 1− p and person 3 plays3A with probability p and 3B with probability 1 − p. If person 1 plays 1A, her expectedpayoff is 3p2 + 4(1 − p)2. If person 1 plays 1B, her expected payoff is 4p2. To find theswitchover probability, we set 3p2 + 4(1 − p)2 = 4p2 and hence get 4(1 − p)2 = p2. Thus4− 8p + 4p2 = p2 and so 3p2 − 8p + 4 = 0. Factoring, we get (3p− 2)(p− 2) = 0 and hencep = 2/3 (p = 2 is not possible because p is a probability). If you do the same thing withpersons 2 and 3, you will get the same answer for p since the game is so symmetric.Hence in the mixed Nash equilibrium, each person plays A with probability 2/3 and B withprobability 1/3.

46. Say that Townsville is deciding how many coal-fired energy plants to build to supplyits energy needs. Some people are more environmentally oriented and thus prefer fewerplants, and some people think that the jobs and electricity that the plants provide are moreimportant. Hence people differ on how many plants they feel are necessary. An opinionpoll is taken asking each person how many plants she or he prefers. The results are that14 percent of the population prefer 0 plants, 16 percent prefer 1 plant, 18 percent prefer 2plants, 6 percent prefer 3 plants, 30 percent prefer 4 plants, and 16 percent prefer 5 plants.

a. Say that there are two candidates running for office, and the only relevant issue is howmany plants to build. Each candidate takes a position on how many plants to build, andthen each voter votes for the candidate which is closest to her own position or “ideal point.”For example, if candidate 1 is for 3 plants and candidate 2 is for 0 plants, a voter who prefers2 plants will vote for candidate 1. If there is a “tie” (if two candidates are equally close toa voter’s ideal point), then half of the votes go to each candidate. For example, if candidate1 is for 2 plants and candidate 2 is for 0 plants, then half of the people who prefer 1 votefor candidate 1 and half vote for candidate 2. Each candidate wants to maximize the totalnumber of votes she gets.

Model this as a strategic form game (the candidates move simultaneously) as in the Downsianmodel. Find the pure strategy Nash equilibrium. Predict what positions the candidates willtake and how many plants the town will build.The game looks like this:

0 1 2 3 4 5

0 50, 50 14, 86 22, 78 30, 70 39, 61 48, 52

1 86, 14 50, 50 30, 70 39, 61 48, 52 51, 49

2 78, 22 70, 30 50, 50 48, 52 51, 49 54, 46

3 70, 30 61, 39 52, 48 50, 50 54, 46 69, 31

4 61, 39 52, 48 51, 49 46, 54 50, 50 84, 16

5 52, 48 49, 51 46, 54 31, 69 16, 84 50, 50

The only pure strategy Nash equilibrium is (3, 3). Both candidates support building 3 plants.Note that the median voter (the 50th percentile voter) favors building 3 plants.

b. Now say that there are three candidates. Is there a pure strategy Nash equilibrium whichis similar to what you found in part a.?If there were three candidates, and all picked 3, then each would get a payoff of 33.3. If oneof them deviated and picked (for example) 4, then the person deviating would get 46, whichis better than 33.3. Hence (3, 3, 3) would not be a Nash equilibrium of the three persongame.

c. Now say that opinions shift. A new poll is taken, and it is found that 4 percent of thepopulation prefer 0 plants, 10 percent prefer 1 plant, 78 percent prefer 2 plants, 2 percentprefer 3 plants, 2 percent prefer 4 plants, and 4 percent prefer 5 plants. Say there are twocandidates. Predict what positions the candidates will take and how many plants the townwill build.Now the game looks like this:

0 1 2 3 4 5

0 50, 50 4, 96 9, 91 14, 86 53, 47 92, 8

1 96, 4 50, 50 14, 86 53, 47 92, 8 93, 7

2 91, 9 86, 14 50, 50 92, 8 93, 7 94, 6

3 86, 14 47, 53 8, 92 50, 50 94, 6 95, 5

4 47, 53 8, 92 7, 93 6, 94 50, 50 96, 4

5 8, 92 7, 93 6, 94 5, 95 4, 96 50, 50

The only pure strategy Nash equilibrium is (2, 2). Both candidates support building 2 plants.Note that the median voter (the 50th percentile voter) favors building 2 plants.

d. Now again say that there are three candidates. Is there a pure strategy Nash equilibriumwhich is similar to what you found in part c.?If there were three candidates, and all picked 2, then each would get a payoff of 33.3. Ifone of them deviated and picked 0, then the person deviating would get 7.33. If a persondeviated to 1, she would get 14. If a person deviated to 3, she would get 8. If a persondeviated to 4, she would get 6.67. If a person deviated to 5, she would get 6. Hence a personcannot gain by deviating. Hence (3, 3, 3) is a Nash equilibrium.

Note that in part a, all three candidates taking the median voter position was not a Nashequilibrium. Here, because there are so many people at the “center” (78 percent prefer 2plants), a candidate cannot gain by taking a more left or right position, even if there arethree candidates.

47. Say that Gotham City is deciding how many skate parks and dog walks to build inthe city. A survey is done and it is found that 40 percent of the population are crankytaxpayers who dislike public expenditures and prefer 0 skate parks and 0 dog walks; 22percent are hard core skate punks who prefer 6 skate parks and 0 dog walks; 30 percent areyuppie golden retriever owners who prefer 0 skate parks and 6 dog walks; and 8 percent areconsensus-minded Buddhists who prefer 2 skate parks and 2 dog walks.Say that there are two candidates running for office who take positions on both issues. As inthe Downsian model, each voter votes for the candidate which is closest to her own positionor “ideal point.” For example, if candidate 1 favors 1 skate park and 1 dog walk, andcandidate 2 favors 4 skate parks and 2 dog walks, then candidate 1 gets 78 percent of thevote (the cranky taxpayers, the yuppies, and the Buddhists) and candidate 2 gets 22 percent(the skate punks). Each candidate wants to maximize the total number of votes she gets.

a. Let’s try to make a prediction in this game by eliminating weakly and strongly dominatedstrategies. First, simplify the game a lot by considering only the following strategies: (0,0),(0,3), (0,6), (1,1), (2,2), (3,0), (3,3), (6,0). Here (3,0) means for example 3 skate parks and0 dog walks. Each of the two candidates thus has eight possible strategies. Write this as astrategic form game and make a prediction by eliminating weakly and strongly dominatedstrategies.The game looks like this (we show only person 1’s payoffs; person 2’s payoff is 100 minusperson 1’s payoff).

(0, 0) (0, 3) (0, 6) (1, 1) (2, 2) (3, 0) (3, 3) (6, 0)

(0, 0) 50 62 70 40 40 70 40 78

(0, 3) 38 50 70 30 30 54 70 78

(0, 6) 30 30 50 30 30 30 30 54

(1, 1) 60 70 70 50 40 78 44 78

(2, 2) 60 70 70 60 50 78 48 78

(3, 0) 30 46 70 22 22 50 62 78

(3, 3) 60 30 70 56 52 38 50 78

(6, 0) 22 22 46 22 22 22 22 50

In this game, the strategies (0,0), (0,6), (1,1), (3,0), (6,0) are weakly or strongly dominatedand we are left with (0,3),(2,2),(3,3). No further elimination is possible.

(0, 3) (2, 2) (3, 3)

(0, 3) 50 30 70

(2, 2) 70 50 48

(3, 3) 30 52 50

b. Now say that most of the yuppies see Richard Gere movies and decide to become Buddhists(and take the Buddhist position of supporting 2 skate parks and 2 dog walks). Now there

are 5 percent yuppies, 33 percent Buddhists, 40 percent cranky taxpayers, and 22 percentskate punks. Again, find all strategies for both candidates which are not weakly dominated.Now we get the following game (again, for simplicity only candidate 1’s payoffs are given).

(0, 0) (0, 3) (0, 6) (1, 1) (2, 2) (3, 0) (3, 3) (6, 0)

(0, 0) 50 62 95 40 40 45 40 78

(0, 3) 38 50 95 5 5 41.5 45 78

(0, 6) 5 5 50 5 5 5 5 41.5

(1, 1) 60 95 95 50 40 78 56.5 78

(2, 2) 60 95 95 60 50 78 73 78

(3, 0) 55 58.5 95 22 22 50 62 78

(3, 3) 60 55 95 43.5 27 38 50 78

(6, 0) 22 22 58.5 22 22 22 22 50

In this game, the weakly or strongly dominated strategies are (0,0), (0, 3), (0, 6), (1,1), (3,0),(3,3), and (6,0). Only (2,2) remains. The shift in the electorate toward central positionscauses the candidates to take more centrist positions.

48. The city council of Asbestosville wants to improve its image by bringing in the Palookav-ille Pirates, a minor league baseball franchise which is currently located in Palookaville.Asbestosville has built a brand new baseball field and is now trying to come up with otherenticements for the Pirates, such as how much cash to give to the team. The city has alreadyagreed to give the team $1 million, but some council members want to give the team moremoney. There are 11 council members. One member does not want to give the team anymore money and prefers to give only $1 million total to the Pirates, one member prefers togive $2 million in total, one member prefers to give $3 million, one member prefers to give$4 million, and so forth; the eleventh council member wants to give the Pirates $11 millionin total. As you can see, the median member of the council wants to give the Pirates $6million.

a. The city council chairperson is a baseball fanatic and wants to give the Pirates $11 millionin total. The chairperson controls the city agenda and thus decides what proposal to bringto the council. When a proposal is brought to the council, all council members simply voteyes or no (the chairperson also votes). If the vote fails, then the policy remains at the statusquo (giving $1 million total). Like in the Downsian model, a council member wants the finalpolicy to be as close to her own “ideal point” as possible. What proposal will the chairpersonmake? How much money will the Pirates receive?We can model this as an extensive form game and find the subgame perfect Nash equilibria,but it is easy enough to simply reason through it. If the chair proposes $6 million for example,council member 4 will vote for it (since 6 is closer to 4 than the status quo, 1, is). For thatmatter, council members 4 through 11 will vote for it. Council member 3 will vote againstit (since the status quo 1 is closer than 6), and so will council members 1 and 2. Hence a $6million proposal will pass by majority. Since the chairperson wants to give as much moneyas possible to the team, she will propose the highest possible amount such that a majoritywill vote for it over the status quo. If the chairperson proposes $10 million, council members6 through 11 will vote for it (6 votes in favor) and council members 1 through 5 will vote

against (5 votes against). Hence this proposal will pass. If the chairperson proposes themore extreme $11 million, council members 7 through 11 will vote for it (5 votes in favor)and council members 1 through 5 will vote against it (5 votes against); council member 6will be indifferent and might vote either way (or “split” her vote evenly among the two).Since council member 6 is “shaky,” we’ll say that the chairperson will propose $10 millionand it will pass for certain. So the Pirates will receive $10 million.

b. Now say that the mayor can veto the city council decision. The mayor’s ideal point isto give the Pirates a total of $2.6 million. If the council’s decision is farther away thanthe status quo from the mayor’s ideal point, then the mayor will veto the council’s decisionand the status quo will be implemented. The sequence of decisions is like this: the councilchairperson first makes a proposal, then the council members vote, and then the mayor canveto. Now what proposal will the chair person make? How much money will the Piratesreceive?Again, we can model this as an extensive form game, but let’s just reason through it. Forthe mayor, the status quo (1) is 1.6 away from his ideal point (2.6). So he will veto anyproposal which is more than 1.6 away (any proposal higher than 4.2). The only proposalswhich he will not veto are 1, 2, 3, and 4. Given this, the council chairperson will propose$4 million, which will pass by a vote of 9 to 2, and will not be vetoed by the mayor. ThePirates will receive $4 million. If we allow the chairperson to propose fractional amounts,the chairperson will propose $4.2 million.

c. Now say that if the mayor vetoes the city council decision, the city council can overridethe veto with two thirds of the council vote (in this case, 8 of the 11 council members). Ifthe council’s proposal is vetoed, then if the council overrides the veto, it can implement itsoriginal proposal. So now the sequence of decisions is like this: the council chairperson firstmakes a proposal, then the council members vote, and then the mayor can veto; if the mayorvetoes, then the council can override. Now what proposal will the chair person make? Howmuch money will the Pirates receive?If eight council members prefer a proposal to the status quo, then the council will overridethe mayor’s veto in favor of the proposal. Note that the $6 million proposal would get eightvotes (council members 4 through 11) in favor, three against. The $7 million proposal willget seven votes in favor (members 5 through 11), three against (members 1 through 3) andmember 4 will be indifferent between the two proposals. An $8 million or higher proposalwill get seven votes or fewer, and hence a proposal of $8 million would not have support oftwo thirds of the council over the status quo. So $6 million is the highest proposal whichwill get two thirds support for certain. A $7 million proposal might get two thirds, but isshaky because of the indifference of member 4. Since the chair prefers the highest possibleproposal, she will propose $6 million, it will pass, and not be vetoed.

49. Say that there are three people and five candidates {a, b, c, d, e}. Say person 1’s order ofpreference (from best to worst) is c, b, e, d, a. Person 2’s order is d, c, a, b, e. Person 3’s orderis e, a, b, d, c.

a. Show that for each candidate, there is an order of voting (an “agenda”) in which thatcandidate wins.The easiest way to do this problem is first figure out which candidate beats which by majority.It’s easy to find that a > b, b > d, b > e, c > a, c > b, c > e, d > a, d > c, e > a, e > d,where a > b means that a beats b by majority rule.There are many possible agendas which make a win. The easiest way to find one is to firstfigure out which candidate a beats (in this case b). Then we figure out which candidate bbeats (say d). Then we figure out which candidate d beats (in this case c). Then we figureout which candidate c beats (in this case e). So we have the sequence a, b, d, c, e.Thus our agenda goes like this: First, vote over a or not. If not, then vote over b or not. Ifnot, vote over d or not. If not, vote between c and e.If the last (fourth) stage is reached, c will be chosen (c beats e by majority). If the thirdstage is reached, d wins (d beats c). If the second stage is reached, b wins (b beats d). Inthe first stage, a wins since it beats b.

b. Say that there are three people and four candidates {a, b, c, d}. Say person 1’s order ofpreference (from best to worst) is c, b, d, a. Person 2’s order is b, a, d, c. Person 3’s order isa, c, d, b. Show that there is no order of voting in which candidate d wins. Why is this?There is no agenda in which d wins simply because d does not beat any other candidateby majority rule. Regardless of the agenda, at some point d will have to be compared withsome other candidate, and when this happens, d will lose.

50. Say that eleven people vote over four candidates {a, b, c, d}. Three people have preferenceorder (from best to worst) b, a, c, d. Three people have preference order b, a, d, c. Three peoplehave preference order a, c, d, b. Two people have preference order a, d, c, b.

a. Is there a candidate which beats all others by majority rule (a Condorcet winner)?It is easy to find that b beats all other candidates by majority rule. Hence b is a Condorcetwinner. Candidate a beats both c and d but is beaten by b.

b. Say that instead of standard majority rule voting, that each person votes 3 points for theirfirst choice, 2 points for their second choice, 1 point for their third choice, and no points fortheir last choice. This procedure is called the “Borda count.” Which candidate wins now?Now a gets 27 points, b gets 18 points, c gets 11 points, and d gets 10 points. Candidate awins by the Borda count.

c. Which procedure (majority rule or Borda count) is more reasonable or more fair in youropinion?One could argue that a is fairer overall because it is everyone’s first or second choice. Can-didate b beats a by majority rule, but for many people (5 out of 11) b is the worst choice.

51. Another system of voting is “approval voting,” in which each voter can place vote for asmany candidates as she wishes: for example, one person might put one vote on candidate a,and another person might put one vote on candidate a and one vote on candidate b. Say wehave approval voting, and each person votes for his top two candidates.

a. Say that candidate a receives more votes by approval voting than the other two candidates.Is it possible for a to not be a Condorcet winner? If so, write down the preference orderingsfor each person which make this possible.Of course, there are many possible examples. The one I thought of has five people and threecandidates. Persons 1, 2, and 3 rank the candidates from best to worst as b, a, c. Persons 4and 5 rank the candidates as a, c, b. Here b beats all other candidates by majority rule andis thus the Condorcet winner.However, under approval voting, candidate a gets 5 votes, candidate b gets 3 votes, andcandidate c gets 2 votes. Thus candidate a wins under approval voting. In this sense, likethe Borda count, approval voting is ”fairer” than majority rule.

b. Say that a is a Condorcet winner. Is it possible for a to receive fewer approval votes thansome other candidate? If so, write down the preference orderings for each person which makethis possible.One can use a similar example. Say persons 1, 2, and 3 rank the candidates from best toworst as a, b, c and persons 4 and 5 rank the candidates as b, c, a. Here a beats all othercandidates by majority rule and is thus the Condorcet winner, but b wins under approvalvoting.

52. Say that there are three people who are deciding over three alternatives, a, b, c. Is itpossible for the Borda count winner to be different from the Condorcet winner? If so, writedown the preference orderings for each person which make this possible.Note that since each person casts 3 Borda points, there are a total of 9 Borda points. Henceit is impossible to be a Borda count winner with just 3 points (either someone else will get4 points or more, or the other two people will get 3 points each, in which case everyone willtie for first place). Thus, to be a Borda count winner, you must receive at least 4 points.If you receive 6 Borda count points, you are everyone’s first choice and hence you are ob-viously the Condorcet winner. If you receive 5 Borda count points, you are the first choiceof two people and the second choice of a third. Since a majority (two people) have you astheir first choice, you must be the Condorcet winner. If you receive 4 Borda count points,you must either be two people’s first choice and one person’s last choice, or one person’s firstchoice and two people’s second choice. If you are two people’s first choice, you must be theCondorcet winner. If you are one person’s first choice and two people’s second choice, theonly way that another candidate can beat you by majority is if that other candidate is twopeople’s first choice. But then that other candidate would receive 4 Borda count points atleast, which must mean you are not a Borda count winner.Hence in this particular example (3 candidates, 3 voters), it is impossible for a Borda countwinner to not be a Condorcet winner.

53. Say that there are three groups in society. Group X’s preferences (from best to worst) area, b, c. Group Y’s preferences (from best to worst) are c, b, a. Group Z’s preferences (frombest to worst) are b, c, a. There are 13 people in group Y and 7 people in group Z. There arex people in group X.

a. Depending on the value of x, what is the Condorcet winner? Is there some value of x suchthat there is no Condorcet winner?I should have said this in the problem, but we assume throughout that the number of peoplein society is odd to avoid ties. Since the total number of people in society is x + 20, assumex is odd.In a vote between a and b, a gets x votes (group X) and b gets 20 votes (groups Y and Z).Similarly, in a vote between a and c, a gets x votes and c gets 20 votes. In a vote betweenb and c, b gets x + 7 votes (groups X and Z) and c gets 13 votes. Thus if x ≥ 21, then abeats both b and c by majority, and a is the Condorcet winner. If x ≤ 19, then a loses toboth b and c, and the only contest is between b and c. If 7 ≤ x ≤ 19, then b beats a and cby majority, and hence b is the Condorcet winner. If x ≤ 5, then c beats a and b and hencec is the Condorcet winner.As long as we assume x is odd (and hence there are no ties), there is always a Condorcetwinner in this example.

b. Say society uses the Borda count system. For what values of x does the outcome of theBorda count differ from what the society would choose if they chose the Condorcet winner?Under the Borda count system, a gets 2x points (each person in X gives a 2 points), b gets27 + x points (each person in X gives b one point, each person in Y gives b one point, andeach person in Z gives b 2 points), and c gets 33 points (each person in Y gives c 2 pointsand each person in Z gives c one point).Thus if 2x > 27 +x and 2x > 33, then a wins. Doing some algebra, this is true when x > 27(or in other words, when x ≥ 29, since x is assumed odd). When x = 27, then a and b tiefor first place in the Borda count. If 7 ≤ x ≤ 25, then b wins in the Borda count becausethen 27 + x > 2x and 27 + x > 33. If x ≤ 5, then c wins in the Borda count.Comparing this with part a. above, we find that if x ≥ 29, then a is both the Condorcetwinner and the Borda count winner. If x = 27, then a is the Condorcet winner but ties withb in the Borda count. If x = 21, 23, 25 then a is the Condorcet winner but b is the Bordacount winner. If 7 ≤ x ≤ 19, then b is the Condorcet winner and the Borda count winner.If x ≤ 5, then c is both the Condorcet winner and the Borda count winner.So in this example, the Condorcet winner and the Borda count winner are different whenx = 21, 23, 25.

c. Say society uses approval voting in which each person votes for her top two candidates.For what values of x does the outcome of this approval voting system differ from what thesociety would choose if they chose the Condorcet winner?When each person votes for her top two candidates, a gets x votes, b gets 20 + x votes, andc gets 20 votes. Notice that everyone now votes for b, since it is everyone’s first or secondchoice. Hence b always wins.Comparing this with part a. above, we find that when x ≥ 21, a is the Condorcet winnerwhile b is the approval voting winner. When 7 ≤ x ≤ 19, b is the Condorcet winner and the

approval voting winner. When x ≤ 5, c is the Condorcet winner but b is the appoval votingwinner.Hence the approval voting winner differs from the Condorcet winner when x ≥ 21 or x ≤ 5.

54. Say that there are three people deciding by majority rule over four candidates a, b, c, d.Person 1’s preferences (from best to worst) are a, b, c, d. Person 2’s preferences (from best toworst) are c, d, b, a. Person 3’s preferences (from best to worst) are d, a, c, b. Consider votingagendas in which people vote on candidates sequentially.We can write a > b, a > c, c > b, c > d, d > a, d > b, where a > b means a beats b bymajority rule.

a. Is there an agenda in which they decide on a? If there is, show it. If not, explain why.How about this agenda: first a or not; if not, c or not; if not, d or b. Going backward fromthe end: d beats b, so if the last vote is reached, d will win. Given this, c will win on thesecond vote. Given this, a will win on the first vote.

b. Is there an agenda in which they decide on b? If there is, show it. If not, explain why.There is no agenda in which b wins because b does not beat any other candidate by majority.

c. Is there an agenda in which they decide on c? If there is, show it. If not, explain why.How about this agenda: first c or not; if not, d or not; if not, a or b. Going backward fromthe end: a beats b, so if the last vote is reached, a will win. Given this, d will win on thesecond vote. Given this, c will win on the first vote.

d. Is there an agenda in which they decide on d? If there is, show it. If not, explain why.How about this agenda: first d or not; if not, a or not; if not, b or c. Going backward fromthe end: c beats b, so if the last vote is reached, c will win. Given this, a will win on thesecond vote. Given this, d will win on the first vote.

55. Say that there are three people deciding by majority rule over eight candidates{a, b, c, d, e, f, g, h}. Person 1’s preferences (from best to worst) are f, c, g, e, b, h, a, d. Person2’s preferences (from best to worst) are a, f, e, g, h, b, d, c. Person 3’s preferences (from bestto worst) are g, a, b, c, h, d, e, f . Consider voting agendas in which people vote on candidatessequentially.

a. Find the top cycle.It is not hard to see that a beats everything but g, f beats everything but a, and g beatseverything but f . In other words, a is beaten only by g, f is beaten only by a, and g isbeaten only by f . Thus the top cycle is a, f, g. No other candidate is in the top cycle. Forexample, if b were in the top cycle, it would have to beat something which beats something,and so forth, which eventually beats a. But the only thing which beats a is g, and the onlything which beats g is f , and b does not beat f or g (or a for that matter).

b. For every candidate in the top cycle, find a voting agenda in which that candidate wins.An agenda which implements a is as follows. First vote on a or not. If a loses, vote on f ornot. If f loses, vote on g or not. Then consider the rest of the candidates in any order.

This agenda works because g would beat any of the candidates other than a or f . Whenthey vote on f , then f is chosen because f beats g. Thus when they vote for a, they votefor a because a beats f .Agendas which implement f and g are similar.

56. Say that we have a threshold model in which there are 5 people. If the total number ofother people who participate is greater or equal to a person’s threshold, the person wantsto participate also. If the total number of other people who are participating is less than aperson’s threshold, the person does not want to participate.

a. Say that one person has threshold 1, two people have threshold 2, and two people havethreshold 4. Find all of the pure strategy Nash equilibria.One Nash equilibrium is (n, n, n, n, n), in which no one participates—since no one partici-pates, no one’s threshold is met. Another is (p, p, p, p, p), in which everyone participates—everyone’s threshold is met (for example, the threshold 4 people see 4 other people partici-pating). Another is (p, p, p, n, n)—the threshold 1 and the two threshold 2 people participatebecause their thresholds are met.

b. Now say that one of the threshold 2 people becomes a threshold 0 person. Find all of thepure strategy Nash equilibria. Does this change guarantee some level of participation?The threshold 0 person participates for sure. Hence the threshold 1 joins in, and then thethreshold 2 person joins in. However, a threshold 4 person does not join in because thereare only three other people participating. So one Nash equilibrium is (p, p, p, n, n). Anotheris (p, p, p, p, p)—if everyone starts off by participating, then everyone wants to continue toparticipate. The outcome (n, n, n, n, n), however, is no longer a Nash equilibrium. Hencethe change guarantees at least some level of participation.

57. Say that you have a group of 50 people who can either buy a color fax machine or notbuy. No one wants to buy a color fax machine if no one else has one (because there wouldbe no one to exchange color faxes with). In fact, each person will buy one only if at least 6other people buy them. Thus each person has a threshold of 6.

a. Find the two pure strategy Nash equilibria.The pure strategy Nash equilibria are (p, . . . , p) (everyone participating) and (n, . . . , n) (noone participating).

b. Now say that you are a sales rep for the color fax machine company. You can offer discountcoupons to potential customers. If you give 1 discount coupon to someone, that decreasestheir threshold by 1. For example, if you give 6 discount coupons to a single person, youcan make that person have threshold 0. If you give 2 discount coupons to a single person,you can make that person have threshold 4. Obviously, you can guarantee that everyone willbuy a color fax machine by giving all 50 people six coupons each, but that would be silly(the company would get no profits). Using the fewest possible number of coupons, how canyou guarantee that everyone will buy a color fax machine?To guarantee that someone will participate for sure, you need to have one person who hasthreshold 0 (you give that person 6 coupons). Given that this person participates, you needto make one person have threshold 1 (you give that person 5 coupons). Similarly, you give 4

coupons to another person to make that person have threshold 2, you give 3 coupons to makeanother person have threshold 3, and so forth. Hence you give out 6 coupons to 1 person, 5coupons to another person, 4, to another person, 3 to another person, 2 to another person,and 1 to another person. So the resulting thresholds are 0, 1, 2, 3, 4, 5, 6, 6, 6, . . . , 6. The firstsix people (with thresholds 0 to 5) will revolt, and then everyone else’s threshold will be metand then everyone else will revolt. You give out a total of 6 + 5 + 4 + 3 + 2 + 1 = 21 coupons.

58. [from Spring 2003 final] Say that we have 10 people. Each person is thinking aboutwhether or not to join a revolt or not. Each person has a threshold: five people havethreshold 4 and five people have threshold 6.

a. Find all pure strategy Nash equilibria of this game.Everyone not revolting is a Nash equilibrium; if no one revolts, no one wants to revolt. Ifsomeone revolts, at least five people must revolt (since the lowest threshold is threshold 4).If at least five people revolt, then all of the threshold 4 people revolt. If all five threshold 4people revolt, this alone is not enough to get the threshold 6 people to revolt. So anotherNash equilibrium is the threshold 4 people revolting and the threshold 6 people not revolting.Say one of the threshold 6 people revolts; then all of the threshold 6 people want to revolt.Hence another Nash equilibrium is everyone revolting.So there are three Nash equilibria: no one revolts, only the threshold 4 people revolt, andeveryone revolts.

b. Say that you have some discount coupons which lower the cost of revolting and hence lowera person’s threshold. For example, if I give 3 coupons to a person with threshold 4, she nowhas threshold 1. If I give 1 coupon to a person with threshold 6, he now has threshold 5.By giving out coupons, I can change the game so that the only Nash equilibrium is one inwhich everyone revolts. I want to do this by giving out the fewest number of coupons. Howdo I distribute the coupons (who gets coupons, and how many does each person get)?To get rid of the Nash equilibrium in which no one revolts, someone must have threshold zero.The “cheapest” way to make someone have threshold zero is to give 4 coupons to a threshold4 person. To make another person revolt for sure, we need to make her have threshold 1;so we can give 3 coupons to a threshold 4 person. Similarly, we make another threshold 4person have threshold 2 by giving him 2 coupons and make another threshold 4 person havethreshold 3 by giving him 1 coupon. Finally, we give a threshold 6 person 1 coupon and makehim a threshold 5 person. So in other words, we started with thresholds 4, 4, 4, 4, 4, 6, 6, 6, 6, 6and we end up with thresholds 0, 1, 2, 3, 4, 5, 6, 6, 6, 6. We give out 4 + 3 + 2 + 1 + 0 + 1 = 11coupons.

c. Now say that you can give tickets which raise the cost of revolting and hence raise aperson’s threshold. For example, if I give 3 tickets to a person with threshold 4, she nowhas threshold 7. If I give 2 tickets to a person with threshold 6, he now has threshold 8.By giving out tickets, I can change the game so that the only Nash equilibrium is one inwhich no one revolts. I want to do this by giving out the fewest number of tickets. How doI distribute the tickets (who gets tickets, and how many does each person get)?To eliminate the equilibrium in which everyone revolts, there must be at least one personwith threshold 10. Given that this person never revolts, you can make another person

never revolt by making them have threshold 9, and so forth. So we start with thresholds4, 4, 4, 4, 4, 6, 6, 6, 6, 6 and end up with thresholds 4, 4, 4, 4, 5, 6, 7, 8, 9, 10. We give out 1+0+1 + 2 + 3 + 4 = 11 coupons.

59. Say that there are four people: Alicia, Betsy, Carlos, and Davis. Each can choose whetherto wear platform sandals or not. Alicia is very fashion-forward and will wear them even ifno one else wears them; in fact, if more than one other person wears them, she won’t wearthem anymore because she hates being part of a crowd. Betsy also likes fashion but is notas cutting-edge: she will wear them if at least one other person wears them, but like Alicia,hates being “part of the crowd” and will not wear them if more than two other people wearthem. Carlos thinks of himself as hip, but is kind of slow on the uptake and will wear them ifat least two others wear them. Still, Carlos has at least some fashion pride and will not wearthem if everyone else wears them. Finally, Davis gets his fashion tips from the JC Penneycatalog and will wear them if everyone else wears them.

a. Say that at the beginning, no one wears platform sandals (they have just hit the market).Show that at first the sales of platform sandals steadily grow, but eventually sales “cycle”between high and low in a never-ending “fashion cycle.”Starting from (n, n, n, n) (no one buys), first Alicia only wears platform sandals. Hence weget (p, n, n, n). Given this, Betsy jumps in and we get (p, p, n, n). Given this, Carlos jumps inand we get (p, p, p, n). Now Davis jumps in but Alicia drops out because they are becomingtoo popular; hence we get (n, p, p, p). Now Davis drops out and we get (n, p, p, n). NowCarlos drops out and we get (n, p, n, n). Now Alicia jumps back in and Betsy drops out, andwe get (p, n, n, n). Then the cycle starts all over again.

b. Are there any pure strategy Nash equilibria of this game?There are sixteen possibilities to consider (p, p, p, p), (p, n, n, n), and so forth. To avoidhaving to go through all sixteen, we first realize that the only way that Davis participatesis if everyone else participates. It is easy to show that (p, p, p, p) is not a Nash equilibrium(Alicia wants to drop out). Hence we can assume that Davis never participates. Given this,Carlos participates only if Alicia and Betsy participate: (p, p, p, n). It is easy to see thatthis is not a Nash equilibrium (Alicia wants to drop out). Hence we can assume that Carlosdoes not participate. So there are only four possibilities: (p, p, n, n), (p, n, n, n), (n, p, n, n),and (n, n, n, n). It is easy to check that none of these is a Nash equilibrium. So there are nopure strategy Nash equilibria of this game.

60. Say that there are three men A, B, and C and three women X, Y, and Z. Each of thesesix people is considering matching up with a member of the opposite sex. Each person wouldrather be matched with someone than not have a partner at all. Man A prefers woman Xbest, woman Y next, and woman Z least. Man B prefers woman Y best, woman Z next, andwoman X least. Man C prefers woman Y best, woman X next, and woman Z least. WomanX prefers man B best, man A next, and man C least. Woman Y prefers man A best, manB next, and man C least. Woman Z prefers man A best, man C next, and man B least.

a. Say that man A is matched with woman X, man B is matched with woman Y, and manC is matched with woman Z. Is this matching stable?

Yes, this match is stable. Men A and B get their first choice. Man C would like to call X orY, but both X and Y like their current partners over C. Woman X would like to call B, butB is attached to his first choice. Women Y and Z would like to call A, but A is attached tohis first choice.

b. Write down all possible matchings and determine which of them are stable and which arenot stable.(AX, BY, CZ) is stable, as mentioned above. (AX, BZ, CY) is not stable—B and Y wouldlike to dump their current partners and join with each other. (AY, BX, CZ) is stable. (AY,BZ, CX) is not stable—A and X would like to get together. (AZ, BX, CY) is not stable—Aand Y would like to get together. (AZ, BY, CX) is not stable—A and Y would like to gettogether.

c. Among the set of stable matchings, which matching is most preferred by the men? Amongthe set of stable matchings, which matching is most preferred by the women?All of the men prefer the stable matching (AX, BY, CZ) at least as much as the other stablematching, (AY, BX, CZ). All of the women prefer the stable matching (AY, BX, CZ) at leastas much as the other stable matching, (AX, BY, CZ).

61. Say that there are four men, A, B, C, and D and four women W, X, Y, and Z. Eachof these eight people is considering matching up with a member of the opposite sex. Eachperson would rather be matched with someone than not have a partner at all. Man A preferswoman W best, X next, Y, next, and Z least. Man B’s preference ordering (from best toworst) is Y, X, W, Z. Man C’s preference ordering is Y, Z, X, W. Man D’s ordering is Y,Z, X, W. Woman W’s preference ordering (from best to worst) is D, C, B, A. Woman X’spreference ordering is C, B, A, D. Woman Y’s ordering is D, A, C, B. Woman Z’s orderingis B, C, D, A.

a. Among the set of stable matchings, which matching is most preferred by the men? Amongthe set of stable matchings, which matching is most preferred by the women?To find the stable matching most preferred by the men, we use the “men ask” algorithm.First everyone asks their first choice: A asks W, B asks Y, C asks Y, and D asks Y. WomanY likes D best out of all the men who ask her, and thus B and C are rejected. So B and Cask their second choices: B asks X and C asks Z. As it stands, A is matched with W, B withX, C with Z, and D with Y, and there are no women who are asked by more than one man.Hence (AW, BX, CZ, DY) is a stable matching.Now let’s use the “women ask” algorithm. First W asks D, X asks C, Y asks D, and Z asksB. Man D is asked by both W and Y, and so D rejects W and pairs up with Y. So W thenasks C. Now C is asked by two people (W and X), and C chooses to pair up with X. So thecurrent matches are XC, YD, and ZB.Since W is rejected again, W asks B. Now B is asked by W and Z, and B chooses to pair upwith W. So the current matches are now WB, XC, and YD. Since Z is rejected, she asks C.Now C is asked by X and Z, and C chooses Z. Now the current matches are WB, YD, andZC. Since X is rejected, she asks B. Now B is asked by both W and X, and B chooses X.Now the current matches are XB, YD, and ZC. Now W asks A. Since no man is now askedby more than one woman, the matching (WA, XB, YD, ZC) is stable.

Note that this matching is the same as what we obtained in the “men ask” algorithm. Sincethe best matching for the women is the same as the best matching for the men, in this casethis matching is the only stable matching.

62. Say that Person 1 and Person 2 each decide whether to go to the auto racing match orthe ballet. If they go to different places, both get utility zero. If they both go to the autoracing match, person 1 gets a utility of $4 and person 2 gets a utility of $1. If they both goto the ballet, person 1 gets a utility of $1 and person 2 gets a utility of $4. This is a “battleof the sexes” game. Now say that before playing this battle of the sexes game, person 1 caneither burn $2 or not burn it. Person 2 can see whether person 1 burns the money or not.

a. Represent this as a strategic form game (each person has four strategies) and find all (purestrategy) Nash equilibria.

b. Show that by iteratively eliminating both strongly and weakly dominated strategies, onecan predict that person 1 does not burn the money and that both go to the auto racingmatch.

63. Say that you and I are playing a game in which we both simultaneously yell out either aor b. If we say the same letter, then you get $1 and I get nothing. If we say different letters,than I get $1 and you get nothing.

a. Model this as a game and find the Nash equilibrium.

b. Now say that I just had my wisdom teeth pulled out and my mouth is still quite numb.So it’s more difficult for me to say b: when I say b, I have to pay the cost r, where r > 0.Now we play the same game above (my payoffs have changed but yours have not). Modelthis as a game and find the Nash equilibrium.

c. How does my expected utility in the Nash equilibrium change as r changes? What happenswhen r = 1? When r > 1?

64. Say that person 1 and person 2 are playing a drinking game which goes like this. Thereare m beers in the refrigerator. Person 1 goes first by drinking either 1 or 2 beers. Thenperson 2 can drink either 1 or 2 beers. Then person 1 can drink either 1 or 2 beers, and soforth. In other words, when it is a person’s turn to drink, she can drink either 1 or 2 beers.Whoever drinks the last beer wins the game. Winning the game yields a payoff of 1 andlosing yields a payoff of 0. However, there is an additional feature to the game: there is a“magic number” x (which is greater than 0 and less than m). If after your turn, there areexactly x beers left, then you lose the game and also have to go out and buy more beer; thishas a payoff of −3 for the loser and a payoff of 1 for the winner.

a. Say that m = 6 and x = 4. Model this as an extensive form game and find a subgameperfect Nash equilibrium.

b. Say that m = 6 and x = 3. Model this as an extensive form game and find a subgameperfect Nash equilibrium.

c. Now let m and x be any number. Find a subgame perfect Nash equilibrium. For whatvalues of m and x can person 1 guarantee a win? For what values of m and x can person 2guarantee a win?

65. [from Spring 2002 final] There are three people who each simultaneously choose the letterA or the letter B. If a person chooses a letter which no one else chooses, then he gets a payoffof 4; if a person chooses a letter which some other person chooses, then he gets a payoff of 0.The only exception is if everyone chooses the letter A, in which case everyone gets a payoffof 3.

a. Model this as a strategic form game and find all pure-strategy Nash equilibria.

b. There exists a mixed-strategy Nash equilibrium in which each person plays A with prob-ability p and B with probability 1− p. Find p.

66. [from Spring 2003 final] Say that we have 10 people. Each person is thinking aboutwhether or not to join a revolt or not. Each person has a threshold: five people havethreshold 4 and five people have threshold 6.

a. Find all pure strategy Nash equilibria of this game.

b. Say that you have some discount coupons which lower the cost of revolting and hence lowera person’s threshold. For example, if I give 3 coupons to a person with threshold 4, she nowhas threshold 1. If I give 1 coupon to a person with threshold 6, he now has threshold 5.By giving out coupons, I can change the game so that the only Nash equilibrium is one inwhich everyone revolts. I want to do this by giving out the fewest number of coupons. Howdo I distribute the coupons (who gets coupons, and how many does each person get)?

c. Now say that you can give tickets which raise the cost of revolting and hence raise aperson’s threshold. For example, if I give 3 tickets to a person with threshold 4, she nowhas threshold 7. If I give 2 tickets to a person with threshold 6, he now has threshold 8.By giving out tickets, I can change the game so that the only Nash equilibrium is one inwhich no one revolts. I want to do this by giving out the fewest number of tickets. How doI distribute the tickets (who gets tickets, and how many does each person get)?

67. Say that we have three men, A, B, and C, and three women X, Y, and Z. Each personhas preferences about which member of the opposite sex they would like to be matched with.For example, say woman X likes A the best, B second best, and C worst. Assume that thereare no ties (i.e. it cannot be that woman Y likes A the best and B and C are tied for worst).Note that there are six possible matchings.

a. Is it possible for people’s preferences to be such that there exists exactly one stablematching? If so, write down the preferences which make this possible. If not, explain whynot.

b. Is it possible for people’s preferences to be such that there exists exactly two stablematchings? If so, write down the preferences which make this possible. If not, explain whynot.

c. Is it possible for people’s preferences to be such that there exists exactly three stablematchings? If so, write down the preferences which make this possible. If not, explain whynot.

d. Is it possible for people’s preferences to be such that there exists exactly four stablematchings? If so, write down the preferences which make this possible. If not, explain whynot.

68. Say that there are five people choosing among three candidates x, y, z. Persons 1 and 2’spreference ordering from best to worst is x, z, y. Persons 3 and 4 have preference orderingy, z, x. Person 5 has preference ordering z, x, y.

a. Say that they make their decision using a runoff procedure: first everyone votes for theirfirst choice, and the two alternatives which get the most votes go to a runoff. In the runoff,each person votes for one of these two candidates, and whoever gets the most votes in therunoff wins. Which candidate wins in this procedure?

b. Is the candidate who wins the runoff the Condorcet winner? Is there a Condorcet winner?

69. Say that there are three people choosing among five candidates s, t, w, x, y. Person1’s preference ordering from best to worst is w, x, t, y, s. Person 2’s preference ordering isy, w, x, t, s. Person 3’s preference ordering is s, x, t, y, w.

a. Say that people decide using majority rule according to some agenda. For example, oneagenda might be to vote on w first; if w loses, then vote on x; if x loses, then vote on s; if sloses, then vote on t versus y. Can you find an agenda in which t is chosen?

b. What is the top cycle? Remember that the candidates in the top cycle are those whichwin given some suitably chosen agenda. Candidates not in the top cycle are those which arenever chosen regardless of the agenda.

70. Say that the town of Mollusk Beach is deciding how many oil refineries to build. Fortypercent of voters prefer no refineries, 36 percent prefer 1 refinery, and 24 percent prefer tworefineries. Say that there are two candidates A and B, and each has to take a position onthis issue. Given their positions, each voter will vote for the candidate whose position isclosest to their own (if there are two candidates who are equally far away, assume that thevote is split equally among the two candidates). Each candidate wants to maximize the totalnumber of votes she receives.

a. Say that candidates A and B choose their positions simultaneously. Which positions willthey take?

b. Say now that candidate A can choose her position first and then candidate B chooses herposition. Which positions will they take?

c. Now say that there are three candidates, A, B, and C. Say that candidate A chooses herposition first, then candidate B, and then finally candidate C. Which positions will theytake?

Homework answers IEE1149 Summer 2015 - Social Sciences2. Read \Which Price is Right?" by Charles...

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